hdu 4289 最小割,分拆点为边

题目链接:http://acm.zju.edu.cn/onlinejudge/showProblem.do?problemCode=2609

#include <cstdio>
#include <cmath>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <queue>
#include <vector>

#define maxn 450
#define maxe 100000 
using namespace std;

const int INF = 0x3f3f3f;

struct Edge{
    int from,to,cap,flow;
    int next;
};

struct Dinic{
    int s,t;
    int head[maxn];
    int cur[maxn];
    Edge edges[maxe];
    int d[maxn];
    bool vis[maxn];
    int cnt;
    
       void init(){
         memset(head,-1,sizeof(head));
         cnt = 0;      
    }
    void addedge(int from,int to,int cap){
        edges[cnt].from = from; edges[cnt].to = to;   edges[cnt].cap = cap; 
        edges[cnt].flow = 0   ; edges[cnt].next = head[from];  head[from] = cnt++;
        edges[cnt].from = to  ; edges[cnt].to = from; edges[cnt].cap = 0; 
        edges[cnt].flow = 0   ; edges[cnt].next = head[to];  head[to] = cnt++;
    }
    bool bfs(){
        memset(vis,0,sizeof(vis)); 
        queue<int> Q;
        Q.push(s);  
        vis[s] = true;  
        d[s] = 0;
        while(!Q.empty()){
            int u = Q.front();  Q.pop(); 
            for(int i=head[u];i!=-1;i=edges[i].next){
                Edge& e = edges[i];    
                if(!vis[e.to] && e.cap>e.flow){
                    vis[e.to] = true;
                    d[e.to] = d[e.from] + 1;
                    Q.push(e.to);
                }
            }
        }
        return vis[t];
    }
    int dfs(int u,int res){
        if( u == t ||  res == 0)   return res; 
        int flow = 0,f;
        for(int& i=cur[u];i!=-1;i=edges[i].next){   //还不是很理解到cur[]的作用; 
            Edge& e = edges[i];
            if(d[e.to] == d[e.from] + 1 && (f = dfs(e.to,min(res,e.cap-e.flow)))>0){
                e.flow += f;
                edges[i^1].flow -= f;  
                flow += f;
                res -= f;
                if(res == 0) break; 
            } 
        }
        return flow;
    }
    int Maxflow(int S,int T){
        s = S; t = T;
        int flow = 0;
        while(bfs()){
            for(int i=s;i<=t;i++)  cur[i] = head[i];
            flow += dfs(s,INF);   
        }
        return flow;
    }
}solver;

int main()
{
    //if(freopen("input.txt","r",stdin)== NULL)  {printf("Error\n"); exit(0);}
    
    int N,M,S,T;
    while(scanf("%d%d",&N,&M) == 2){
        scanf("%d%d",&S,&T);
        solver.init();
        int s,t;
        s = 0;  t = 2*N + 1;
        solver.addedge(s,S,INF);
        solver.addedge(T+N,t,INF);  //
        for(int i=1;i<=N;i++){
            int cost;
            scanf("%d",&cost);
            solver.addedge(i,i+N,cost);
        }
        for(int i=1;i<=M;i++){  
            int a,b;
            scanf("%d %d",&a,&b);
            solver.addedge(b+N,a,INF);
            solver.addedge(a+N,b,INF);
        }
        printf("%d\n",solver.Maxflow(s,t));
    }
}
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posted @ 2013-07-31 13:52  等待最好的两个人  阅读(124)  评论(0编辑  收藏  举报