POJ1273

Problem : Drainage Ditches
Descripition : 有一天下雨了，然后从1到N排水，有M条通道，每条通道有一个最大排水量，现在问你最多能排多少水。
Solution : 裸最大网络流。
Code(C++) :

#include <stdio.h>#include <string.h>#include <iostream>#include <string>#include <queue>#define MIN(a,b) ((a)>(b)? (b):(a))using namespace std;const int SIZE=400+50;const int INF=0x3f3f3f3f;struct Node{    int u,v;    int cap;    Node(){}    Node(int U,int V,int Cap):    u(U),v(V),cap(Cap){}};int src,des;vector<int> G[SIZE];Node e[SIZE*SIZE/2];int d[SIZE];int cur[SIZE];int n;int top;int M,N;void addedge(int u,int v,int cap){    G[u].push_back(top);    e[top++]=Node(u,v,cap);    G[v].push_back(top);    e[top++]=Node(v,u,0);}void build(){       int i,j;    for(i=0;i<SIZE;i++)        G[i].clear();    top=0;    n=N;    src=0;    des=N+1;    int x,y,c;    for(i=1;i<=M;i++){        scanf("%d%d%d",&x,&y,&c);        addedge(x,y,c);    }    addedge(src,1,INF);    addedge(n,des,INF);}bool bfs(int src,int des){    memset(d,-1,sizeof(d));    queue<int> que;    que.push(src);    d[src]=0;    while(!que.empty()){        int now=que.front();        que.pop();        for(int i=0;i<G[now].size();i++){            Node &tmp=e[G[now].at(i)];            if(tmp.cap>0&&d[tmp.v]==-1)                d[tmp.v]=d[now]+1,                que.push(tmp.v);        }    }    return d[des]>=0;}int dfs(int t,int sum,int des){    if(t==des||!sum)        return sum;    int flow=0,f;    for(int &i=cur[t];i<G[t].size();i++){        Node &tmp=e[G[t].at(i)];        if(d[tmp.v]==d[t]+1&&(f=dfs(tmp.v,MIN(sum,tmp.cap),des))>0){            tmp.cap-=f;            e[G[t].at(i)^1].cap+=f;            sum-=f;            flow+=f;            if(!sum)                break;        }    }    return flow;}int DINIC(int src,int des){    int sum=0;    while(bfs(src,des)){        memset(cur,0,sizeof(cur));        sum+=dfs(src,INF,des);    }    return sum;}void work(){    build();    printf("%d\n",DINIC(src,des));}int main(){    //freopen("in.data","r",stdin);    while(~scanf("%d%d",&M,&N))        work();    return 0;}