【HDOJ】4183 Pahom on Water

就是一个网络流。red结点容量为2,查看最大流量是否大于等于2。对于条件2,把边反向加入建图。条件1,边正向加入建图。

  1 /* 4183 */
  2 #include <iostream>
  3 #include <string>
  4 #include <map>
  5 #include <queue>
  6 #include <set>
  7 #include <stack>
  8 #include <vector>
  9 #include <deque>
 10 #include <algorithm>
 11 #include <cstdio>
 12 #include <cmath>
 13 #include <ctime>
 14 #include <cstring>
 15 #include <climits>
 16 #include <cctype>
 17 #include <cassert>
 18 #include <functional>
 19 #include <iterator>
 20 #include <iomanip>
 21 using namespace std;
 22 //#pragma comment(linker,"/STACK:102400000,1024000")
 23 
 24 #define sti                set<int>
 25 #define stpii            set<pair<int, int> >
 26 #define mpii            map<int,int>
 27 #define vi                vector<int>
 28 #define pii                pair<int,int>
 29 #define vpii            vector<pair<int,int> >
 30 #define rep(i, a, n)     for (int i=a;i<n;++i)
 31 #define per(i, a, n)     for (int i=n-1;i>=a;--i)
 32 #define clr                clear
 33 #define pb                 push_back
 34 #define mp                 make_pair
 35 #define fir                first
 36 #define sec                second
 37 #define all(x)             (x).begin(),(x).end()
 38 #define SZ(x)             ((int)(x).size())
 39 #define lson            l, mid, rt<<1
 40 #define rson            mid+1, r, rt<<1|1
 41 
 42 typedef struct {
 43     int u, v, c, f;
 44 } Edge_t;
 45 
 46 const int maxn = 605;
 47 Edge_t E[maxn*maxn];
 48 vi vc[maxn];
 49 double F[maxn];
 50 int X[maxn], Y[maxn], R[maxn];
 51 int pre[maxn], a[maxn], ID[maxn];
 52 int n, m;
 53 
 54 void addEdge(int u, int v, int c) {
 55     E[m].u = u;
 56     E[m].v = v;
 57     E[m].c = c;
 58     E[m].f = 0;
 59     
 60     E[m+1].u = v;
 61     E[m+1].v = u;
 62     E[m+1].c = 0;
 63     E[m+1].f = 0;
 64     
 65     vc[u].pb(m);
 66     vc[v].pb(m+1);
 67     
 68     m += 2;
 69 }
 70 
 71 bool judge(int i, int j) {
 72     if (F[i] >= F[j])
 73         return false;
 74     if ((R[i]+R[j])*(R[i]+R[j]) > (X[i]-X[j])*(X[i]-X[j])+(Y[i]-Y[j])*(Y[i]-Y[j]))
 75         return true;
 76     return false;
 77 }
 78 
 79 bool bfs(int s, int t) {
 80     int u, v;
 81     int sz, k;
 82     queue<int> Q;
 83     
 84     memset(a, 0, sizeof(a));
 85     Q.push(s);
 86     a[s] = INT_MAX;
 87     pre[s] = s;
 88     
 89     while (!Q.empty()) {
 90         u = Q.front();
 91         Q.pop();
 92         sz = SZ(vc[u]);
 93         rep(i, 0, sz) {
 94             k = vc[u][i];
 95             v = E[k].v;
 96             if (!a[v] && E[k].f<E[k].c) {
 97                 a[v] = min(a[u], E[k].c-E[k].f);
 98                 pre[v] = u;
 99                 ID[v] = k;
100                 Q.push(v);
101             }
102         }
103     }
104     
105     return a[t]==0;
106 }
107 
108 int EK(int s, int t) {
109     int ret = 0, tmp;
110     int u, v, k;
111     
112     while (1) {
113         if (bfs(s, t))
114             break;
115         
116         tmp = a[t];
117         for (v=t,u=pre[v]; v!=s; v=u,u=pre[v]) {
118             k = ID[v];
119             E[k].f += tmp;
120             E[k^1].f -= tmp;
121         }
122         ret += tmp;
123     }
124     
125     return ret;
126 }
127 
128 int main() {
129     ios::sync_with_stdio(false);
130     #ifndef ONLINE_JUDGE
131         freopen("data.in", "r", stdin);
132         freopen("data.out", "w", stdout);
133     #endif
134     
135     int case_n, n_;
136     int s, t;
137     int ans;
138     
139     scanf("%d", &case_n);
140     while (case_n--) {
141         scanf("%d", &n_);
142         n = n_ + n_;
143         m = 0;
144         
145         rep(i, 0, n)
146             vc[i].clr();
147         
148         rep(i, 0, n_) {
149             scanf("%lf %d %d %d", &F[i], &X[i], &Y[i], &R[i]);
150             if (F[i] == 400.0)
151                 s = i;
152             else if (F[i] == 789.0)
153                 t = i;
154         }
155         
156         if (judge(s, t)) {
157             puts("Game is VALID");
158             continue;
159         }
160         
161         rep(i, 0, n_) {
162             if (i!=s && i!=t)
163                 addEdge(i<<1, i<<1|1, 1);
164             rep(j, 0, i) {
165                 if (judge(i, j))
166                     addEdge(i<<1|1, j<<1, 1);
167                 else if (judge(j, i))
168                     addEdge(j<<1|1, i<<1, 1);
169             }
170         }
171         
172         addEdge(s<<1, s<<1|1, 2);
173         
174         ans = EK(s<<1, t<<1);
175         
176         if (ans >= 2)
177             puts("Game is VALID");
178         else
179             puts("Game is NOT VALID");
180     }
181     
182     #ifndef ONLINE_JUDGE
183         printf("time = %d.\n", (int)clock());
184     #endif
185     
186     return 0;
187 }

 

posted on 2015-10-09 23:34  Bombe  阅读(198)  评论(0编辑  收藏  举报

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