poj2112Optimal Milking(二分+最大流）

链接

floyd求出牛到机器的最短距离，二分距离，小于当前距离的边容量设为1，求出满容量下的最短距离。

EK算法

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 #include<cmath>
 #include<queue>
 #include<set>
 using namespace std;
 #define N 255
 #define LL long long
 #define INF 0xfffffff
 const double eps = 1e-;
 const double pi = acos(-1.0);
 const double inf = ~0u>>;
 int path[N],flow[N],gh[N][N],st,en;
 int w[N][N];
 int bfs()
 {
     int i;
     memset(path,-,sizeof(path));
     for(i =  ; i <= en ; i++)
     flow[i] = INF;
     queue<int>q;
     q.push();
     while(!q.empty())
     {
         int tk = q.front();
         q.pop();
         if(tk==en)
         break;
         for(i =  ; i <= en ; i++)
         {
             if(path[i]==-&&gh[tk][i])
             {
                 path[i] = tk;
                 flow[i] = min(flow[tk],gh[tk][i]);
                 q.push(i);
             }
         }
     }
     if(path[en]==-)
     return -;
     return flow[en];
 }
 int EK()
 {
     int now,pre,sum=,k;
     while((k=bfs())!=-)
     {
         sum+=k;
         now = en;
         while(now!=st)
         {
             pre = path[now];
             gh[pre][now]-=k;
             gh[now][pre]+=k;
             now = pre;
         }
     }
     return sum;
 }
 int main()
 {
     int k,c,i,g,j,m;
     while(scanf("%d%d%d",&k,&c,&m)!=EOF)
     {
         int n = k+c;
         for(i = ; i <= n+; i++)
             for(j = ;j <= n+ ; j++)
                 w[i][j] = INF;
         for(i = ;i <= n+ ; i++)
             for(j = ; j <= n+ ; j++)
             {
                 scanf("%d",&w[i][j]);
                 if(w[i][j]==) w[i][j] = INF;
                 if(i==j) w[i][i] = ;
                 w[j][i] = w[i][j];
             }
         for(i =  ; i <= n+; i++)
             for(j = ; j <= n+ ;j++)
                 for(g = ;g <= n+; g++)
                 w[j][g] = min(w[j][g],w[j][i]+w[i][g]);
         st = ;
         en = n+;
         int low = ,high = ,mid;
         while(low<=high)
         {
             mid = (low+high)>>;
             memset(gh,,sizeof(gh));
             for(i = ; i <= k+ ; i++)
             gh[i][en] = m;
             for(i = k+;  i <= n+; i++)
             gh[st][i] = ;
             for(i = k+; i <= n+; i++)
                 for(j = ; j <= k+ ; j++)
                 if(w[i][j]<=mid)
                 gh[i][j] = w[i][j];
             //cout<<EK()<<endl;
             if(EK()==c)
             high = mid-;
             else
             low =mid+;
         }
         cout<<low<<endl;
     }
     return ;
 }