poj1273（最大流）

Description

Every time it rains on Farmer John's fields, a pond forms over Bessie's favorite clover patch. This means that the clover is covered by water for awhile and takes quite a long time to regrow. Thus, Farmer John has built a set of drainage ditches so that Bessie's clover patch is never covered in water. Instead, the water is drained to a nearby stream. Being an ace engineer, Farmer John has also installed regulators at the beginning of each ditch, so he can control at what rate water flows into that ditch. 
Farmer John knows not only how many gallons of water each ditch can transport per minute but also the exact layout of the ditches, which feed out of the pond and into each other and stream in a potentially complex network. 
Given all this information, determine the maximum rate at which water can be transported out of the pond and into the stream. For any given ditch, water flows in only one direction, but there might be a way that water can flow in a circle. 

Input

The input includes several cases. For each case, the first line contains two space-separated integers, N (0 <= N <= 200) and M (2 <= M <= 200). N is the number of ditches that Farmer John has dug. M is the number of intersections points for those ditches. Intersection 1 is the pond. Intersection point M is the stream. Each of the following N lines contains three integers, Si, Ei, and Ci. Si and Ei (1 <= Si, Ei <= M) designate the intersections between which this ditch flows. Water will flow through this ditch from Si to Ei. Ci (0 <= Ci <= 10,000,000) is the maximum rate at which water will flow through the ditch.

Output

For each case, output a single integer, the maximum rate at which water may emptied from the pond.

Sample Input

5 41 2 401 4 202 4 202 3 303 4 10

Sample Output

50

最大流的精髓在于怎样建图。

这题是模板题，直接上模板就搞定了：

#include<stdio.h>#include<iostream>using   namespace std;const   int oo=1e9;/**oo 表示无穷大*/ const  int mm=111111111; /**mm 表示边的最大数量，记住要是原图的两倍，在加边的时候都是双向的*/ const  int mn=999; /**mn 表示点的最大数量*/ int node,src,dest,edge; /**node 表示节点数，src 表示源点，dest 表示汇点，edge 统计边数*/ int ver[mm],flow[mm],next[mm]; /**ver 边指向的节点，flow 边的容量 ，next 链表的下一条边*/ int head[mn],work[mn],dis[mn],q[mn];void prepare(int _node, int _src,int _dest){ node=_node,src=_src,dest=_dest; for(int i=0;i<node;++i)head[i]=-1; edge=0; } /**增加一条 u 到 v 容量为 c 的边*/ void addedge( int u,  int v,  int c) { ver[edge]=v,flow[edge]=c,next[edge]=head[u],head[u]=edge++; ver[edge]=u,flow[edge]=0,next[edge]=head[v],head[v]=edge++; } /**广搜计算出每个点与源点的最短距离，如果不能到达汇点说明算法结束*/ bool Dinic_bfs() { int i,u,v,l,r=0; for(i=0;i<node;++i)dis[i]=-1; dis[q[r++]=src]=0; for(l=0;l<r;++l) for(i=head[u=q[l]];i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]<0) {/**这条边必须有剩余容量*/ dis[q[r++]=v]=dis[u]+1; if(v==dest)  return 1; } return 0; } /**寻找可行流的增广路算法，按节点的距离来找，加快速度*/ int Dinic_dfs(  int u, int exp) { if(u==dest)  return exp; /**work 是临时链表头，这里用 i 引用它，这样寻找过的边不再寻找*/ for(  int &i=work[u],v,tmp;i>=0;i=next[i]) if(flow[i]&&dis[v=ver[i]]==dis[u]+1&&(tmp=Dinic_dfs(v,min(exp,flow[i])))>0) { flow[i]-=tmp; flow[i^1]+=tmp; /**正反向边容量改变*/ return tmp;} return 0; }int Dinic_flow() { int i,ret=0,delta; while(Dinic_bfs()) { for(i=0;i<node;++i)work[i]=head[i]; while(delta=Dinic_dfs(src,oo))ret+=delta; } return ret; } //此处省略最大流模板,详细见最大流模板int main(){int n,m,u,v,c;while(~scanf("%d%d",&n,&m)){ prepare(m+1,1,m);while(n--){scanf("%d%d%d",&u,&v,&c);addedge(u,v,c);} printf("%d\n",Dinic_flow()); } return 0;}