poj1273:http://poj.org/problem?id=1273
题意：现在有m个池塘(从1到m开始编号,1为源点,m为汇点),及n条水渠,给出这n条水渠所连接的点和所能流过的最大流量，
题解：裸的网络流。
1 #include<iostream> 2 #include<cstring> 3 #include<algorithm> 4 #include<cstdio> 5 #include<queue> 6 #define INF 100000000 7 using namespace std; 8 const int N=405; 9 const int M=1000000; 10 struct Node{ 11 int v; 12 int f; 13 int next; 14 }edge[M]; 15 int n,m,u,v,w,cnt,sx,ex; 16 int head[N],pre[N]; 17 void init(){ 18 cnt=0; 19 memset(head,-1,sizeof(head)); 20 } 21 void add(int u,int v,int w){ 22 edge[cnt].v=v; 23 edge[cnt].f=w; 24 edge[cnt].next=head[u]; 25 head[u]=cnt++; 26 edge[cnt].f=0; 27 edge[cnt].v=u; 28 edge[cnt].next=head[v]; 29 head[v]=cnt++; 30 } 31 bool BFS(){ 32 memset(pre,0,sizeof(pre)); 33 pre[sx]=1; 34 queue<int>Q; 35 Q.push(sx); 36 while(!Q.empty()){ 37 int d=Q.front(); 38 Q.pop(); 39 for(int i=head[d];i!=-1;i=edge[i].next ){ 40 if(edge[i].f&&!pre[edge[i].v]){ 41 pre[edge[i].v]=pre[d]+1; 42 Q.push(edge[i].v); 43 } 44 } 45 } 46 return pre[ex]>0; 47 } 48 int dinic(int flow,int ps){ 49 int f=flow; 50 if(ps==ex)return f; 51 for(int i=head[ps];i!=-1;i=edge[i].next){ 52 if(edge[i].f&&pre[edge[i].v]==pre[ps]+1){ 53 int a=edge[i].f; 54 int t=dinic(min(a,flow),edge[i].v); 55 edge[i].f-=t; 56 edge[i^1].f+=t; 57 flow-=t; 58 if(flow<=0)break; 59 } 60 61 } 62 if(f-flow<=0)pre[ps]=-1; 63 return f-flow; 64 } 65 int solve(){ 66 int sum=0; 67 while(BFS()) 68 sum+=dinic(INF,sx); 69 return sum; 70 } 71 int main() { 72 while(~scanf("%d%d",&n,&m)) { 73 init(); 74 for(int i=1;i<=n; i++) { 75 scanf("%d%d%d",&u,&v,&w); 76 add(u,v,w); 77 } 78 sx=1;ex=m; 79 printf("%d ",solve()); 80 } 81 return 0; 82 }