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
1 #include <iostream> 2 #include<cstdio> 3 #include<cstring> 4 #include<algorithm> 5 #include<stdlib.h> 6 #include<vector> 7 #include<cmath> 8 #include<queue> 9 #include<set> 10 using namespace std; 11 #define N 255 12 #define LL long long 13 #define INF 0xfffffff 14 const double eps = 1e-8; 15 const double pi = acos(-1.0); 16 const double inf = ~0u>>2; 17 int path[N],flow[N],gh[N][N],st,en; 18 int w[N][N]; 19 int bfs() 20 { 21 int i; 22 memset(path,-1,sizeof(path)); 23 for(i = 1 ; i <= en ; i++) 24 flow[i] = INF; 25 queue<int>q; 26 q.push(1); 27 while(!q.empty()) 28 { 29 int tk = q.front(); 30 q.pop(); 31 if(tk==en) 32 break; 33 for(i = 1 ; i <= en ; i++) 34 { 35 if(path[i]==-1&&gh[tk][i]) 36 { 37 path[i] = tk; 38 flow[i] = min(flow[tk],gh[tk][i]); 39 q.push(i); 40 } 41 } 42 } 43 if(path[en]==-1) 44 return -1; 45 return flow[en]; 46 } 47 int EK() 48 { 49 int now,pre,sum=0,k; 50 while((k=bfs())!=-1) 51 { 52 sum+=k; 53 now = en; 54 while(now!=st) 55 { 56 pre = path[now]; 57 gh[pre][now]-=k; 58 gh[now][pre]+=k; 59 now = pre; 60 } 61 } 62 return sum; 63 } 64 int main() 65 { 66 int k,c,i,g,j,m; 67 while(scanf("%d%d%d",&k,&c,&m)!=EOF) 68 { 69 int n = k+c; 70 for(i = 2; i <= n+1; i++) 71 for(j = 2;j <= n+1 ; j++) 72 w[i][j] = INF; 73 for(i = 2;i <= n+1 ; i++) 74 for(j = 2; j <= n+1 ; j++) 75 { 76 scanf("%d",&w[i][j]); 77 if(w[i][j]==0) w[i][j] = INF; 78 if(i==j) w[i][i] = 0; 79 w[j][i] = w[i][j]; 80 } 81 for(i = 2 ; i <= n+1; i++) 82 for(j = 2; j <= n+1 ;j++) 83 for(g = 2;g <= n+1; g++) 84 w[j][g] = min(w[j][g],w[j][i]+w[i][g]); 85 st = 1; 86 en = n+2; 87 int low = 0,high = 10000,mid; 88 while(low<=high) 89 { 90 mid = (low+high)>>1; 91 memset(gh,0,sizeof(gh)); 92 for(i = 2; i <= k+1 ; i++) 93 gh[i][en] = m; 94 for(i = k+2; i <= n+1; i++) 95 gh[st][i] = 1; 96 for(i = k+2; i <= n+1; i++) 97 for(j = 2; j <= k+1 ; j++) 98 if(w[i][j]<=mid) 99 gh[i][j] = w[i][j]; 100 //cout<<EK()<<endl; 101 if(EK()==c) 102 high = mid-1; 103 else 104 low =mid+1; 105 } 106 cout<<low<<endl; 107 } 108 return 0; 109 }