POJ2195(最小费用最大流)

题意：二分图最小权匹配。

构图：S向左边的点连容量1，费用0的边，右边的点向T连容量1，费用0的边，点之间连容量1，费用为边权的边。最小费用最大流。

注意:

1.100*100的方格中有10000个点,数组要开大;

2.n==0 && m==0 时结束。

View Code 
/*Source Code
Problem: 2195  User: HEU_daoguang
Memory: 1172K  Time: 94MS
Language: G++  Result: Accepted
Source Code
*/
#include <iostream>
#include <stdio.h>
#include <queue>
#include <math.h>
#include <string.h>
using namespace std;
#define V 6005
#define E 10010000
#define inf 999999999
int n,m;
char map[102][102];
int hp[V][2],mp[V][2];

int vis[V];
int dist[V];
int pre[V];

struct Edge{
    int u,v,c,cost,next;
}edge[E];
int head[V],cnt;
void init(){
    cnt=0;
    memset(head,-1,sizeof(head));
}

void addedge(int u,int v,int c,int cost){
    edge[cnt].u=u;edge[cnt].v=v;edge[cnt].cost=cost;
    edge[cnt].c=c;edge[cnt].next=head[u];head[u]=cnt++;

    edge[cnt].u=v;edge[cnt].v=u;edge[cnt].cost=-cost;
    edge[cnt].c=0;edge[cnt].next=head[v];head[v]=cnt++;
}

bool spfa(int begin,int end){
    int u,v;
    queue<int> q;

    for(int i=0;i<=end+2;i++){
        pre[i]=-1;
        vis[i]=0;
        dist[i]=inf;
    }
    vis[begin]=1;
    dist[begin]=0;
    q.push(begin);

    while(!q.empty()){

        u=q.front();
        q.pop();
        vis[u]=0;

        for(int i=head[u];i!=-1;i=edge[i].next){
            if(edge[i].c>0){
                v=edge[i].v;
                if(dist[v]>dist[u]+edge[i].cost){
                    dist[v]=dist[u]+edge[i].cost;
                    pre[v]=i;
                    if(!vis[v]){
                        vis[v]=true;
                        q.push(v);
                    }
                }
            }
        }

    }

    return dist[end]!=inf;
}

int MCMF(int begin,int end){
    int ans=0,flow;
    int flow_sum=0;

    while(spfa(begin,end)){

        flow=inf;
        for(int i=pre[end];i!=-1;i=pre[edge[i].u])
            if(edge[i].c<flow)
                flow=edge[i].c;
        for(int i=pre[end];i!=-1;i=pre[edge[i].u]){
            edge[i].c-=flow;
            edge[i^1].c+=flow;
        }
        ans+=dist[end]*flow;
        flow_sum+=flow;

    }
    return ans;
}

int main()
{
    //freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF){
        if(n==0 && m==0) break;
        for(int i=0;i<n;i++){
            scanf("%s",map[i]);
        }
        int hcnt=1,mcnt=1;
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++){
                if(map[i][j]=='H'){
                    hp[hcnt][0]=i;
                    hp[hcnt][1]=j;
                    hcnt++;
                }
                if(map[i][j]=='m'){
                    mp[mcnt][0]=i;
                    mp[mcnt][1]=j;
                    mcnt++;
                }
            }
        hcnt--;
        mcnt--;
        init();
        for(int i=1;i<=hcnt;i++){
            addedge(0,i,1,0);
            //addedge(i,0,1,0);
        }
        for(int j=1;j<=mcnt;j++){
            addedge(hcnt+j,hcnt+mcnt+1,1,0);
            //addedge(hcnt+mcnt+1,hcnt+j,1,0);
        }
        for(int i=1;i<=hcnt;i++)
            for(int j=1;j<=mcnt;j++){
                addedge(i,hcnt+j,1,fabs(hp[i][0]-mp[j][0])+fabs(hp[i][1]-mp[j][1]));
                //addedge(hcnt+j,i,1,fabs(hp[i][0]-mp[j][0])+fabs(hp[i][1]-mp[j][1]));
            }
        int res=MCMF(0,hcnt+mcnt+1);
        printf("%d\n",res);
    }
    return 0;
}
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*/