Optimal Milking（POJ2112+二分+Dinic）

题目链接：http://poj.org/problem?id=2112

题目：

题意：有k台挤奶机，c头奶牛，每台挤奶机每天最多生产m的奶，给你每个物品到其他物品的距离（除了物品到自己本省的距离为0外，两者之间没有路线直接到达也为0，此时需要将距离处理为inf），问跑最远距离的奶牛要跑多远。

思路：先用floyd将各物品间的最短距离求出来，再二分跑最远距离的奶牛走的距离x，将小于x的两者之间进行连边，流量为1，引入一个超级源点和超级汇点，我们采用方向建图的方式，让超级源点与挤奶机连边，且流量为m，奶牛与超级汇点连边，流量为1，跑一边Dinic，如果最大流为c那么ub变成mid-1，否则lb为mid+1。

代码实现如下：

  1 #include <set>
  2 #include <map>
  3 #include <queue>
  4 #include <stack>
  5 #include <cmath>
  6 #include <bitset>
  7 #include <cstdio>
  8 #include <string>
  9 #include <vector>
 10 #include <cstdlib>
 11 #include <cstring>
 12 #include <iostream>
 13 #include <algorithm>
 14 using namespace std;
 15 
 16 typedef long long ll;
 17 typedef pair<ll, ll> pll;
 18 typedef pair<ll, int> pli;
 19 typedef pair<int, ll> pil;;
 20 typedef pair<int, int> pii;
 21 typedef unsigned long long ull;
 22 
 23 #define lson i<<1
 24 #define rson i<<1|1
 25 #define bug printf("*********\n");
 26 #define FIN freopen("D://code//in.txt", "r", stdin);
 27 #define debug(x) cout<<"["<<x<<"]" <<endl;
 28 #define IO ios::sync_with_stdio(false),cin.tie(0);
 29 
 30 const double eps = 1e-8;
 31 const int mod = 10007;
 32 const int maxn = 250 + 7;
 33 const double pi = acos(-1);
 34 const int inf = 0x3f3f3f3f;
 35 const ll INF = 0x3f3f3f3f3f3f3f;
 36 
 37 int k, c, m, s, e, num, tot, maxflow;
 38 int mp[maxn][maxn], head[maxn], d[maxn];
 39 
 40 queue<int> q;
 41 
 42 struct edge {
 43     int v, w, next;
 44 }ed[maxn*maxn];
 45 
 46 void addedge(int u, int v, int w) {
 47     ed[tot].v = v;
 48     ed[tot].w = w;
 49     ed[tot].next = head[u];
 50     head[u] = tot++;
 51     ed[tot].v = u;
 52     ed[tot].w = 0;
 53     ed[tot].next = head[v];
 54     head[v] = tot++;
 55 }
 56 
 57 bool bfs() {
 58     memset(d, 0, sizeof(d));
 59     while(!q.empty()) q.pop();
 60     q.push(s);
 61     d[s] = 1;
 62     while(!q.empty()) {
 63         int x = q.front(); q.pop();
 64         for(int i = head[x]; ~i; i = ed[i].next) {
 65             int y = ed[i].v;
 66             if(ed[i].w && !d[y]) {
 67                 q.push(y);
 68                 d[y] = d[x] + 1;
 69                 if(y == e) return 1;
 70             }
 71         }
 72     }
 73     return 0;
 74 }
 75 
 76 int dinic(int x, int flow) {
 77     if(x == e) return flow;
 78     int rest = flow, k;
 79     for(int i = head[x]; ~i && rest; i = ed[i].next) {
 80         int y = ed[i].v;
 81         if(ed[i].w && d[y] == d[x] + 1) {
 82             k = dinic(y, min(rest, ed[i].w));
 83             if(!k) d[y] = 0;
 84             ed[i].w -= k;
 85             ed[i^1].w += k;
 86             rest -= k;
 87         }
 88     }
 89     return flow - rest;
 90 }
 91 
 92 bool check(int x) {
 93     tot = 0;
 94     memset(head, -1, sizeof(head));
 95     for(int i = 1; i <= k; i++) {
 96         for(int j = k + 1; j <= num; j++) {
 97             if(mp[i][j] <= x) {
 98                 addedge(i, j, 1);
 99             }
100         }
101     }
102     for(int i = 1; i <= k; i++) {
103         addedge(s, i, m);
104     }
105     for(int i = k + 1; i <= num; i++) {
106         addedge(i, e, 1);
107     }
108     int flow = 0;
109     maxflow = 0;
110     while(bfs()) {
111         while((flow = dinic(s, inf))) maxflow += flow;
112     }
113     return maxflow == c;
114 }
115 
116 int main() {
117     //FIN;
118     scanf("%d%d%d", &k, &c, &m);
119     num = k + c;
120     for(int i = 1; i <= num; i++) {
121         for(int j = 1; j <= num; j++) {
122             scanf("%d", &mp[i][j]);
123             if(i != j && mp[i][j] == 0) {
124                 mp[i][j] = inf;
125             }
126         }
127     }
128     for(int k = 1; k <= num; k++) {
129         for(int i = 1; i <= num; i++) {
130             for(int j = 1; j <= num; j++) {
131                 if(mp[i][j] > mp[i][k] + mp[k][j]) {
132                     mp[i][j] = mp[i][k] + mp[k][j];
133                 }
134             }
135         }
136     }
137     s = 0, e = num + 1;
138     int ub = inf, lb = 0, mid;
139     while(ub >= lb) {
140         mid = (ub + lb) >> 1;
141         if(check(mid)) ub = mid - 1;
142         else lb = mid + 1;
143     }
144     printf("%d\n", lb);
145     return 0;
146 }

 

posted @ 2018-08-03 18:56  Dillonh  阅读(186)  评论(0)  编辑  收藏  举报