POJ1273 Drainage Ditches 【最大流】

题目地址

Drainage Ditches
Time Limit: 1000MS Memory Limit: 10000K

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 4
1 2 40
1 4 20
2 4 20
2 3 30
3 4 10

Sample Output
50

裸最大流问题。
人生第一次码最大流，居然还可以1A。
使用的算法为Ford-Fulkerson算法。
说白了就是增广路求最大流。
60+的代码，意外的短。
本来以为这种图论问题随随便便就得100+的Orz
上代码做备忘。

#include<cstdio>#include<cstring>#include<vector>#define INF 2147483647using namespace std;struct edge{    int s,t;    int f,inv;};edge E[405];vector<int> v[405];bool vis[405];int n;int dfs(int p,int maxf){    if(p==n)return maxf;    if(vis[p])return -1;    vis[p]=true;    for(int i=0;i<v[p].size();i++)    {        edge &e=E[v[p][i]];        if(e.f==0)continue;        int pp=dfs(e.t,maxf>e.f?e.f:maxf);        if(pp==-1)continue;        e.f-=pp;        E[e.inv].f+=pp;        return pp;    }    return -1;}int solve(){    int ans=0;    while(1)    {        memset(vis,0,sizeof(vis));        int p=dfs(1,INF);        if(p==-1)break;        ans+=p;    }    return ans;}int main(){    int m;    while(scanf("%d%d",&m,&n)==2)    {        for(int i=1;i<=n;i++)v[i].clear();        for(int i=0;i<m;i++)        {            scanf("%d%d%d",&E[i].s,&E[i].t,&E[i].f);            E[i].inv=i+m;            E[i+m].s=E[i].t;            E[i+m].t=E[i].s;            E[i+m].f=0;            E[i+m].inv=i;            v[E[i].s].push_back(i);            v[E[i].t].push_back(i+m);        }        int ans=solve();        printf("%d\n",ans);    }}