最近网络流AC题目总结（水题版）

1.【模板】网络最大流

这个题目是模板啊，没什么好总结的，只要会EK或Dinic或Isap就可以了，我不会Isap啊，代码什么的在这里点击 我是链接

2. [USACO4.2]草地排水Drainage Ditches

这个题依然是模板啊，m是点，n是边，跑1到n的最短路就A了啊

  1 // luogu-judger-enable-o2
  2 #include <map>
  3 #include <set>
  4 #include <cmath>
  5 #include <ctime>
  6 #include <queue>
  7 #include <stack>
  8 #include <vector>
  9 #include <bitset>
 10 #include <cstdio>
 11 #include <cctype>
 12 #include <string>
 13 #include <cstring>
 14 #include <cassert>
 15 #include <climits>
 16 #include <cstdlib>
 17 #include <iostream>
 18 #include <algorithm>
 19 #include <functional>
 20 using namespace std ;
 21 
 22 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 23 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 24 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 25 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 26 #define clr(a) memset(a,0,sizeof(a))
 27 #define Sort(a, len) sort(a + 1, a + len + 1)
 28 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 29 #define ass(a, sum) memset(a, sum, sizeof(a))
 30 
 31 #define ull unsigned long long
 32 #define ll long long
 33 #define ls ((rt) << 1)
 34 #define rs ((rt) << 1 | 1)
 35 #define mp make_pair
 36 #define pb push_back
 37 #define fi first
 38 #define se second
 39 #define endl '\n'
 40 #define Pii pair<int, int>
 41 
 42 const int N = 100010 ;
 43 const int iinf = INT_MAX/2 ;
 44 const ll linf = LLONG_MAX/2 ;
 45 const int MOD = 1e9+7 ;
 46 
 47 inline int read(){
 48     int X = 0,w = 0 ;
 49     char ch = 0;
 50     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 51     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 52     return w ? -X : X;
 53 }
 54 
 55 void write(int x){
 56      if(x < 0) putchar('-'),x = -x;
 57      if(x > 9) write(x / 10);
 58      putchar(x%10 + '0');
 59 }
 60 
 61 void print(int x) {
 62     cout << x << endl ;
 63     exit(0) ;
 64 }
 65 
 66 struct edge {
 67     int to, nxt, w ;
 68 } e[N << 1];
 69 
 70 int head[N], dep[N] ;
 71 int n, m, s, t, top = 1 ;
 72 
 73 void add(int a, int b, int w) {
 74     e[++top] = (edge) {b, head[a], w} ;
 75     head[a] = top ;
 76 }
 77 
 78 bool bfs() { //分层图 
 79     queue <int> q ;
 80     q.push(s) ;
 81     clr(dep) ;
 82     dep[s] = 1 ;
 83     while (!q.empty()) {
 84         int now = q.front() ;
 85         q.pop() ;
 86         for (int i = head[now]; i; i = e[i].nxt) {
 87             int to = e[i].to ;
 88             if (e[i].w && !dep[to]) {
 89                 dep[to] = dep[now] + 1 ;
 90                 q.push(to) ;
 91             }
 92         }
 93     }
 94     if (!dep[t]) return 0 ;
 95     else return 1 ;
 96 }
 97 
 98 int dfs(int rt, int dis) {
 99     if (rt == t) return dis ;
100     for (int i = head[rt]; i; i = e[i].nxt) {
101         int to = e[i].to ;
102         if (dep[to] == dep[rt] + 1 && e[i].w) {
103             int p = dfs(to, min(dis, e[i].w)) ;
104             if (p) {
105                 e[i].w -= p ;
106                 e[i ^ 1].w += p ;
107                 return p ;
108             }
109         }
110     } 
111     return 0 ;
112 }
113 
114 int Dinic() {
115     int ans = 0 ;
116     while (bfs()) {
117         while (int d = dfs(s, iinf)) ans += d ;
118     }
119     return ans ;
120 }
121 
122 int main() {
123     scanf("%d%d", &m, &n) ;
124     s = 1, t = n ;
125     for (int i = 1; i <= m; i++) {
126         int a, b, c ;
127         scanf("%d%d%d", &a, &b, &c) ;
128         add(a, b, c) ;
129         add(b, a, 0) ;
130     }
131     printf("%d\n", Dinic()) ;
132 }
View Code

3. 酒店之王

这个题目是我第一个正经的网络流题，而且是一个二分图最大匹配。。

我们把人放在房间和菜中间，建一个二分图，图的示意如下：

图中每条边的边权为1，之后再跑网络流（Dinic）即可。

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 5000 ;
 42 const int iinf = INT_MAX/2 ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 int head[N], dep[N], a[N], pre[N] ;
 66 int top = 1, n, p, q, s, t, ans ;
 67 
 68 struct edge {
 69     int to, nxt, w ;
 70 } e[N << 1];
 71 
 72 void add(int a, int b, int w) {
 73     e[++top] = (edge) {b, head[a], w} ;
 74     head[a] = top ;
 75 }
 76 
 77 bool bfs(){
 78     queue <int> q ;
 79     clr(a) ;
 80     a[s] = 0x3f3f3f3f ;
 81     q.push(s) ;
 82     while(!q.empty()){
 83         int now = q.front() ;
 84         q.pop() ;
 85         for(int i=head[now];i;i=e[i].nxt){
 86             int to = e[i].to ;
 87             if(!e[i].w) continue ;
 88             if(a[to]) continue ;
 89             a[to] = min(a[now], e[i].w) ;
 90             pre[to] = i ;
 91             q.push(to) ;
 92         }
 93     }
 94     if(!a[t]) return 0 ;
 95     ans += a[t] ;
 96     return 1 ;
 97 }
 98 
 99 void work(){
100     for (int m = t; pre[m]; m = e[pre[m]^1].to) {
101         e[pre[m]].w -= a[t] ;
102         e[pre[m]^1].w += a[t] ;
103     }
104 }
105 
106 void dinic() {
107     while (bfs()) work() ;
108 }
109 
110 int main() {
111     scanf("%d%d%d", &n, &p, &q) ;
112     s = 2 * n + p + q + 1, t = s + 1 ;
113     for (int i = 1; i <= n; i++)
114     for (int j = 1; j <= p; j++) {
115         int x ;
116         scanf("%d", &x) ;
117         if (x ^ 1) continue ;
118         add(j, p + i, 1) ;
119         add(p + i, j, 0) ;
120     }
121     for (int i = 1; i <= n; i++)
122     for (int j = 1; j <= q; j++) {
123         int x ;
124         scanf("%d", &x) ;
125         if (x ^ 1) continue ;
126         add(p + n + q + i, p + n + j, 1) ;
127         add(p + n + j, p + n + q + i, 0) ;
128     }
129     for (int i = 1; i <= n; i++) {
130         add(p + i, p + n + q + i, 1) ;
131         add(p + n + q + i, p + i, 0) ;
132     }
133     for (int i = 1; i <= p; i++) {
134         add(s, i, 1) ;
135         add(i, s, 0) ;
136     }
137     for (int i = p + n + 1; i <= n + p + q; i++){
138         add(i, t, 1) ;
139         add(t, i, 0) ;
140     }
141     dinic() ;
142     printf("%d\n", ans) ;
143 }
View Code

4.【模板】最小费用最大流

切了前一个紫题，感觉很有成就感，于是又去学新知识，感觉蛮简单的，就是把bfs改成spfa，但始终有一个疑问：能不能用dijkstra跑呢？

建边要从0或2这种细节就不用说了吧

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 100100 ;
 42 const int iinf = INT_MAX ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 int top = 1, n, m, s, t, flow, costs ;
 66 int head[N], dis[N], vis[N] ;
 67 
 68 struct edge {
 69     int to, nxt, w, f ;
 70 } e[N << 1];
 71 
 72 void add(int a, int b, int w, int f) {
 73     e[++top] = (edge) {b, head[a], w, f} ;
 74     head[a] = top ;
 75 }
 76 
 77 struct fuckyou {
 78     int x, id ;
 79 } pre[N];
 80 
 81 bool spfa() {
 82     clr(vis);clr(pre);
 83     ass(dis,0x3f3f3f3f) ;
 84     dis[s] = 0 ;
 85     queue <int> q ;
 86     q.push(s) ;
 87     while (!q.empty()) {
 88         int now = q.front() ;
 89         vis[now] = 0 ;
 90         q.pop() ;
 91         for (int i = head[now]; i; i = e[i].nxt) {
 92             int to = e[i].to ;
 93             if (e[i].w && dis[to] > dis[now] + e[i].f) {
 94                 dis[to] = dis[now] + e[i].f ;
 95                 pre[to] = (fuckyou) {now, i} ;
 96                 if (!vis[to]) {
 97                     q.push(to) ;
 98                     vis[to] = 1 ;
 99                 }
100             }
101         } 
102     } 
103     return dis[t] != 0x3f3f3f3f ;
104 }
105 
106 void ek() {
107     while (spfa()) {
108     //    for (int i = 1; i <= n; i++) cout << dis[i] << " " ;
109     //    cout << endl ; 
110         int mi = iinf ;
111         for (int i = t; i != s; i = pre[i].x) mi = min(mi, e[pre[i].id].w) ;
112         for (int i = t; i != s; i = pre[i].x) {
113             e[pre[i].id].w -= mi ;
114             e[pre[i].id ^ 1].w += mi ; 
115         }
116         flow += mi ;
117         costs += mi * dis[t] ;
118     } 
119 } 
120 
121 int main() {
122     scanf("%d%d%d%d", &n, &m, &s, &t) ;
123     for (int i = 1; i <= m; i++) {
124         int a, b, w, f ;
125         scanf("%d%d%d%d", &a, &b, &w, &f) ;
126         add(a, b, w, f) ;
127         add(b, a, 0, -f) ;
128     }
129 //    cout << "ok\n" ;
130     ek() ;
131     printf("%d %d\n", flow, costs) ;
132 }
View Code

5. [USACO09JAN]全流Total Flow

早上一起来，发现模板全忘光，赶紧打一个模板题。

从A到Z跑网络流即可，是个傻逼题

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 500 ;
 42 const int iinf = INT_MAX/2 ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 int b[300], head[N], dep[N] ;
 66 int top = 1, n = 0, m, s, t, ans ;
 67 
 68 struct edge {
 69     int to, nxt, w ;
 70 } e[N << 1] ;
 71 
 72 void add(int a, int b, int w) {
 73     e[++top] = (edge) {b, head[a], w} ;
 74     head[a] = top ;
 75 }
 76 
 77 bool bfs() {
 78     queue <int> q ;
 79     clr(dep) ;
 80     dep[s] = 1 ;
 81     q.push(s) ;
 82     while (!q.empty()) {
 83         int now = q.front() ;
 84         q.pop() ;
 85         for (int i = head[now]; i; i = e[i].nxt) {
 86             int to = e[i].to ;
 87             if (!dep[to] && e[i].w) {
 88                 dep[to] = dep[now] + 1 ;
 89                 q.push(to) ;
 90             } 
 91         }
 92     }
 93     if (dep[t]) return 1 ;
 94     return 0 ;
 95 }
 96 
 97 int dfs(int rt, int flow) {
 98     if (rt == t) return flow ;
 99     for (int i = head[rt]; i; i = e[i].nxt) {
100         int to = e[i].to ;
101         if (e[i].w && dep[to] == dep[rt] + 1) {
102             int p = dfs(to, min(flow, e[i].w)) ;
103             if (p) {
104                 e[i].w -= p ;
105                 e[i ^ 1].w += p ;
106                 return p ;
107             }
108         }
109     }
110     return 0 ;
111 }
112 
113 int dinic() {
114     while (bfs()) {
115         while (int tmp = dfs(s, iinf)) ans += tmp ;
116     }
117     return ans ;
118 }
119 
120 int main() {
121     cin >> m ;
122     for (int i = 1; i <= m; i++) {
123         char A, B ;
124         int C ;
125         cin >> A >> B >> C ;
126         add(A, B, C) ;
127         add(B, A, 0) ;
128     }
129     s = 'A', t = 'Z' ;
130     printf("%d\n", dinic()) ;
131 }
View Code

6. [CQOI2014]危桥

这个题目也是很水的。

如果这个桥是危桥，建dis为2的边，否则为∞

然后把往返an次和bn次看成走2*an次和2*bn次，由源点向a1和b1连容量分别为2*an和2*bn的边，

再由a2和b2向汇点连容量分别为2*an和2*bn的边，判断最大流是否等于2(an+bn)。

注意坑点：从a1发出的流量有可能跑到b2处。所以还要再建一遍图，源点向a1,b2建边，

a2,b1向汇点建边，如果两次最大流都等于2(an+bn)，那么有解。否则无解。

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int iinf = INT_MAX/2 ;
 42 const ll linf = LLONG_MAX/2 ;
 43 const int MOD = 1e9+7 ;
 44 
 45 inline int read(){
 46     int X = 0,w = 0 ;
 47     char ch = 0;
 48     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 49     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 50     return w ? -X : X;
 51 }
 52 
 53 void write(int x){
 54      if(x < 0) putchar('-'),x = -x;
 55      if(x > 9) write(x / 10);
 56      putchar(x%10 + '0');
 57 }
 58 
 59 
 60 inline char get() {
 61     char c; while ((c = getchar()) != 'X' && c != 'O' && c != 'N');
 62     return c;
 63 }
 64 
 65 const int N = 2e4 + 5, M = 55, INF = 0x3f3f3f3f;
 66 int n, A1, A2, An, B1, B2, Bn, ecnt, nxt[N], adj[N], go[N], cap[N],
 67 S, T, len, que[N], lev[N];
 68 char s[M][M];
 69 
 70 void add_edge(int u, int v, int w) {
 71     nxt[++ecnt] = adj[u]; adj[u] = ecnt; go[ecnt] = v; cap[ecnt] = w;
 72     nxt[++ecnt] = adj[v]; adj[v] = ecnt; go[ecnt] = u; cap[ecnt] = 0;
 73 }
 74 
 75 bool bfs() {
 76     int i; memset(lev, -1, sizeof(lev));
 77     lev[que[len = 1] = S] = 0;
 78     for (i = 1; i <= len; i++) {
 79         int u = que[i];
 80         for (int e = adj[u], v; e; e = nxt[e])
 81             if (cap[e] > 0 && lev[v = go[e]] == -1) {
 82                 lev[que[++len] = v] = lev[u] + 1;
 83                 if (v == T) return 1;
 84             }
 85     }
 86     return 0;
 87 }
 88 
 89 int dinic(int u, int flow) {
 90     if (u == T) return flow;
 91     int res = 0, delta;
 92     for (int e = adj[u], v; e; e = nxt[e])
 93         if (cap[e] > 0 && lev[u] < lev[v = go[e]]) {
 94             delta = dinic(v, min(cap[e], flow - res));
 95             if (delta) {
 96                 cap[e] -= delta; cap[e ^ 1] += delta;
 97                 res += delta; if (res == flow) break;
 98             }
 99         }
100     if (res != flow) lev[u] = -1;
101     return res;
102 }
103 
104 int solve() {
105     int ans = 0;
106     while (bfs()) ans += dinic(S, INF);
107     return ans;
108 }
109 
110 void work() {
111     int i, j, res1, res2;
112     A1 = read() + 2; A2 = read() + 2; An = read();
113     B1 = read() + 2; B2 = read() + 2; Bn = read();
114     for (i = 1; i <= n; i++) for (j = 1; j <= n; j++)
115         s[i][j] = get(); S = 1; T = n + 2;
116     ecnt = 1; memset(adj, 0, sizeof(adj));
117     for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) {
118         if (i == j) continue;
119         if (s[i][j] == 'O') add_edge(i + 1, j + 1, 2);
120         else if (s[i][j] == 'N') add_edge(j + 1, i + 1, INF);
121     }
122     add_edge(S, A1, An << 1); add_edge(S, B1, Bn << 1);
123     add_edge(A2, T, An << 1); add_edge(B2, T, Bn << 1);
124     res1 = solve();
125     ecnt = 1; memset(adj, 0, sizeof(adj));
126     for (i = 1; i <= n; i++) for (j = 1; j <= n; j++) {
127         if (i == j) continue;
128         if (s[i][j] == 'O') add_edge(i + 1, j + 1, 2);
129         else if (s[i][j] == 'N') add_edge(j + 1, i + 1, INF);
130     }
131     add_edge(S, A1, An << 1); add_edge(S, B2, Bn << 1);
132     add_edge(A2, T, An << 1); add_edge(B1, T, Bn << 1);
133     res2 = solve();
134     puts(res1 == (An + Bn << 1) && res2 == (An + Bn << 1) ? "Yes" : "No");
135 }
136 
137 int main() {
138     while (~scanf("%d", &n)) work();
139 }
View Code

7. [USACO4.2]完美的牛栏The Perfect Stall

绿题！非常水！

二分图匹配，直接跑！

s向所有奶牛连边，奶牛向对应牛栏连边，牛栏向t连边，就过了

水啊。。

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 500010 ;
 42 const int iinf = INT_MAX/2 ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 int top = 1, n, m, s, t ;
 66 int head[N], dep[N] ;
 67 
 68 struct edge {
 69     int to, nxt, w ;
 70 } e[N << 1] ;
 71 
 72 void add(int a, int b, int w) {
 73     e[++top] = (edge) {b, head[a], w} ;
 74     head[a] = top ;
 75 }
 76 
 77 bool bfs() {
 78     queue <int> q ; q.push(s) ;
 79     memset(dep, 0, sizeof(dep)) ;
 80     dep[s] = 1 ;
 81     while (!q.empty()) {
 82         int now = q.front() ;
 83         q.pop() ;
 84         for (int i = head[now]; i; i = e[i].nxt) {
 85             int to = e[i].to ;
 86             if (!dep[to] && e[i].w) {
 87                 dep[to] = dep[now] + 1 ;
 88                 q.push(to) ; 
 89             }
 90         }
 91     }
 92     return dep[t] ? 1 : 0 ; 
 93 }
 94 
 95 int dfs(int rt, int flow) {
 96     if (rt == t) return flow ;
 97     for (int i = head[rt]; i; i = e[i].nxt) {
 98         int to = e[i].to ;
 99         if (e[i].w && dep[to] == dep[rt] + 1) {
100             int p = dfs(to, min(flow, e[i].w)) ;
101             if (p) {
102                 e[i].w -= p ;
103                 e[i ^ 1].w += p ;
104                 return p ;
105             }
106         }
107     }
108     return 0 ;
109 }
110 
111 void dinic() {
112     int ans = 0 ;
113     while (bfs()) {
114         while (int d = dfs(s, 0x3f3f3f3f)) ans += d ;
115     }
116     printf("%d\n", ans) ;
117 }
118 
119 int main() {
120     scanf("%d%d", &n, &m) ;
121     for (int i = 1; i <= n; i++) {
122         int sum, a ;
123         scanf("%d", &sum) ;
124         for (int j = 1; j <= sum; j++) {
125             scanf("%d", &a) ;
126             add(i, a + m, 1) ;
127             add(a + m, i, 0) ;
128         }
129     }
130     s = 0, t = n + m + 1 ;
131     for (int i = 1; i <= n; i++) {
132         add(s, i, 1) ;
133         add(i, s, 0) ;
134     }
135     for (int i = n + 1; i <= n + m; i++) {
136         add(i, t, 1) ;
137         add(t, i, 0) ;
138     }
139     dinic() ;
140 }
View Code

8. 地震逃生

又是一个最大流裸题，直接跑，设答案1位ans1，那么$ans2=\left\lceil\dfrac{x}{ans2}\right\rceil$,特判Orz Ni Jinan Saint Cow!的情况就AC了

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 100010 ;
 42 const int iinf = INT_MAX/2 ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 struct edge {
 66     int to, nxt, w ;
 67 } e[N << 1] ;
 68 
 69 int n, m, x, top = 1, s, t, ans ;
 70 int dep[N], head[N] ;
 71 
 72 void add(int a, int b, int w) {
 73     e[++top] = (edge) {b, head[a], w} ;
 74     head[a] = top ;
 75 }
 76 
 77 bool bfs() {
 78     queue <int> q ;
 79     q.push(s) ;
 80     memset(dep, 0, sizeof(dep)) ;
 81     dep[s] = 1 ;
 82     while (!q.empty()) {
 83         int now = q.front() ;
 84         q.pop() ;
 85         for (int i = head[now]; i; i = e[i].nxt){
 86             int to = e[i].to ;
 87             if (!dep[to] && e[i].w) {
 88                 dep[to] = dep[now] + 1 ;
 89                 q.push(to) ;
 90             }
 91         }
 92     }
 93     if (!dep[t]) return 0 ; 
 94     else return 1 ;
 95 }
 96 
 97 int dfs(int rt, int flow) {
 98     if (rt == t) return flow ;
 99     for (int i = head[rt]; i; i = e[i].nxt) {
100         int to = e[i].to ;
101         if (dep[to] == dep[rt] + 1 && e[i].w) {
102             int p = dfs(to, min(flow, e[i].w)) ;
103             if (p) {
104                 e[i].w -= p ;
105                 e[i ^ 1].w += p ;
106                 return p ;
107             }
108         }
109     }
110     return 0 ;
111 }
112 
113 void dinic(){
114     ans = 0 ;
115     while (bfs()) {
116         while (int d = dfs(s, iinf)) ans += d ;
117     }
118 }
119 
120 int work(int a, int b) {
121     if (a % b != 0) return a / b + 1 ;
122     else return a / b ;
123 }
124 
125 int main(){
126     scanf("%d%d%d", &n, &m, &x) ;
127     for (int i = 1; i <= m; i++) {
128         int a, b, c ;
129         scanf("%d%d%d", &a, &b, &c) ;
130         add(a, b, c) ;
131         add(b, a, 0) ;
132     }
133     s = 1, t = n ;
134     dinic() ;
135     if (ans == 0) printf("Orz Ni Jinan Saint Cow!\n") ;
136     else printf("%d %d\n", ans, work(x, ans)) ;
137 }
View Code

9. 飞行员配对方案问题

没有发现原来这里还有一个水题。。。

直接跑网络流求二分图的模板，第二个匹配方案是这样解决的：

to!=s != t (to^1)!=s!=t && w print to && to ^ 1 

ek好像没有这样的问题。

  1 #include <map>
  2 #include <set>
  3 #include <cmath>
  4 #include <ctime>
  5 #include <queue>
  6 #include <stack>
  7 #include <vector>
  8 #include <bitset>
  9 #include <cstdio>
 10 #include <cctype>
 11 #include <string>
 12 #include <cstring>
 13 #include <cassert>
 14 #include <climits>
 15 #include <cstdlib>
 16 #include <iostream>
 17 #include <algorithm>
 18 #include <functional>
 19 using namespace std ;
 20 
 21 #define rep(i, a, b) for (int (i)=(a);(i)<=(b);(i)++)
 22 #define Rep(i, a, b) for (int (i)=(a)-1;(i)<(b);(i)++)
 23 #define REP(i, a, b) for (int (i)=(a);(i)>=(b);(i)--)
 24 #define reg(i, x) for (int (i)=head[x];(i);i=e[i].next)
 25 #define clr(a) memset(a,0,sizeof(a))
 26 #define Sort(a, len) sort(a + 1, a + len + 1)
 27 #define Sort2(a, len, cmp) sort(a + 1, a + len + 1, cmp)
 28 #define ass(a, sum) memset(a, sum, sizeof(a))
 29 
 30 #define ull unsigned long long
 31 #define ll long long
 32 #define ls ((rt) << 1)
 33 #define rs ((rt) << 1 | 1)
 34 #define mp make_pair
 35 #define pb push_back
 36 #define fi first
 37 #define se second
 38 #define endl '\n'
 39 #define Pii pair<int, int>
 40 
 41 const int N = 100010 ;
 42 const int iinf = INT_MAX/2 ;
 43 const ll linf = LLONG_MAX/2 ;
 44 const int MOD = 1e9+7 ;
 45 
 46 inline int read(){
 47     int X = 0,w = 0 ;
 48     char ch = 0;
 49     while(!isdigit(ch)) {w |= ch == '-';ch = getchar();}
 50     while(isdigit(ch)) X = (X<<3) + (X<<1) + (ch ^ 48),ch = getchar();
 51     return w ? -X : X;
 52 }
 53 
 54 void write(int x){
 55      if(x < 0) putchar('-'),x = -x;
 56      if(x > 9) write(x / 10);
 57      putchar(x%10 + '0');
 58 }
 59 
 60 void print(int x) {
 61     cout << x << endl ;
 62     exit(0) ;
 63 }
 64 
 65 int n, m, s, t, top = 1, ans ;
 66 int head[N], dep[N] ;
 67 
 68 struct edge {
 69     int to, nxt, w ;
 70 } e[N << 1] ;
 71 
 72 void add(int a, int b, int w) {
 73     e[++top] = (edge) {b, head[a], w} ;
 74     head[a] = top ;
 75 }
 76 
 77 bool bfs() {
 78     queue <int> q ;
 79     q.push(s) ;
 80     memset(dep, 0, sizeof(dep)) ;
 81     dep[s] = 1 ;
 82     while (!q.empty()) {
 83         int now = q.front() ;
 84         q.pop() ;
 85         for (int i = head[now]; i; i = e[i].nxt) {
 86             int to = e[i].to ;
 87             if (!dep[to] && e[i].w){
 88                 dep[to] = dep[now] + 1 ;
 89                 q.push(to) ;
 90             }
 91         }
 92     }
 93     if (!dep[t]) return 0 ;
 94     else return 1 ;
 95 }
 96 
 97 int dfs(int rt, int flow) {
 98     if (rt == t) return flow ;
 99     for (int i = head[rt]; i; i = e[i].nxt) {
100         int to = e[i].to ;
101         if (dep[to] == dep[rt] + 1 && e[i].w) {
102             int p = dfs(to, min(flow, e[i].w)) ;
103             if (p) {
104                 e[i].w -= p ;
105                 e[i ^ 1].w += p ;
106                 return p ;
107             }
108         }
109     }
110     return 0 ;
111 }
112 
113 void dinic() {
114     ans = 0 ;
115     while (bfs()) {
116         while (int d = dfs(s, iinf)) ans += d ; 
117     }
118 }
119 
120 int main(){
121     scanf("%d%d", &n, &m) ;
122     int a, b ;
123     while (scanf("%d%d", &a, &b) && a != -1 && b != -1) {
124         add(a, b, 1) ;
125         add(b, a, 0) ;
126     }
127     s = 0, t = m + 1 ;
128     for (int i = 1; i <= n; i++) {
129         add(s, i, 1) ;
130         add(i, s, 0) ;
131     }
132     for (int i = n + 1; i <= m; i++) {
133         add(i, t, 1) ;
134         add(t, i, 0) ;
135     }
136     dinic() ;
137     if (!ans) printf("No Solution!\n") ;
138     else {
139         printf("%d\n", ans) ;
140         for (int i = 2; i <= top; i += 2) 
141         if (e[i].to != s && e[i].to != t && e[i ^ 1].to != s && e[i ^ 1].to != t)
142         if (e[i ^ 1].w) printf("%d %d\n", e[i ^ 1].to, e[i].to) ;
143     }
144 }
View Code

10. 教辅的组成

拆点建边，和酒店之王差不多，不讲了

  1 // luogu-judger-enable-o2
  2 #include<iostream>
  3 #include<cstdio>
  4 #include<cstdlib>
  5 #include<cmath>
  6 #include<cstring>
  7 #include<algorithm>
  8 #include<queue>
  9 using namespace std;
 10 #define REP(i, a, b) for(register int i = (a), i##_end_ = (b); i <= i##_end_; ++ i)
 11 #define DREP(i, a, b) for(register int i = (a), i##_end_ = (b); i >= i##_end_; -- i)
 12 #define mem(a, b) memset((a), b, sizeof(a))
 13 int read()
 14 {
 15     int sum = 0, fg = 1; char c = getchar();
 16     while(c < '0' || c > '9') { if (c == '-') fg = -1; c = getchar(); }
 17     while(c >= '0' && c <= '9') { sum = sum * 10 + c - '0'; c = getchar(); }
 18     return sum * fg;
 19 }
 20 #define inf 0x3f3f3f3f
 21 const int maxn = 1000000;
 22 int e,be[maxn], ne[maxn], to[maxn], c[maxn];
 23 int nb, nex, na,m1,m2;
 24 void add(int x, int y, int z)
 25 {
 26     to[e] = y, ne[e] = be[x], be[x] = e;
 27     c[e] = z, e++;
 28     to[e] = x, ne[e] = be[y], be[y] = e;
 29     c[e] = 0, e++;
 30 }
 31 int d[maxn], end;
 32 bool bfs()
 33 {    
 34     queue<int>q;
 35     memset(d,-1,sizeof(d));
 36     q.push(end),d[end] = 0;
 37     while(!q.empty())
 38     {
 39         int u = q.front(); q.pop();
 40         for(int i = be[u]; i!=-1; i = ne[i])
 41         {
 42             int v = to[i];
 43             if(d[v] == -1 && c[i ^ 1])
 44             {
 45                 d[v] = d[u] + 1;
 46                 q.push(v);
 47             }
 48         }
 49     }
 50     return d[0]!=-1;
 51 }
 52 int dfs(int x,int low)
 53 {
 54     if(x == end || !low)return low;
 55     int ret = 0;
 56     for(int i = be[x]; i!=-1;i = ne[i])
 57     {
 58         int v = to[i];
 59         if(d[v] == d[x] - 1 )
 60         {
 61             int    k = dfs(v,min(low-ret,c[i]));
 62             if(k > 0)
 63             {
 64                 c[i] -= k;
 65                 c[i^1] += k;
 66                 ret+=k;
 67             }
 68         }
 69     }
 70     return ret;
 71 }
 72 int dinic()
 73 {
 74     int ans = 0;
 75     while(bfs())
 76     {
 77         int k = dfs(0,inf);
 78         if(k>0)ans+=k;
 79     }
 80     return ans;
 81 }
 82 int main()
 83 {
 84     memset(be,-1,sizeof(be));    
 85     nb = read(); nex = read(); na = read();
 86     m1 = read();
 87     REP(i,1,m1)
 88     {
 89         int x,y;
 90         x = read(), y = read();
 91         add(y,nex+x,1);
 92     }
 93     m2 = read();
 94     REP(i,1,m2)
 95     {
 96         int x,y;
 97         x = read(), y = read();
 98         add(nex+nb+x,2*nb+nex+y,1);
 99     }
100     end = 2*nb+nex+na+1;
101     REP(i,1,nex)    add(0,i,1);
102     REP(i,1,na) add(2*nb+nex+i,end,1);
103     REP(i,1,nb) add(nex+i,nex+nb+i,1);
104     printf("%d",dinic());
105     return 0;
106 }
View Code

thanks for your reading

posted @ 2018-10-21 21:53  harryhqg  阅读(409)  评论(1)  编辑  收藏  举报