poj 2195 Going Home(最小费用最大流)

题目：http://poj.org/problem?id=2195

有若干个人和若干个房子在一个给定网格中，每人走一个都要一定花费，每个房子只能容纳一人，现要求让所有人进入房子，且总花费最小。

构造一个超级源s和超级汇t，超级源s与U中所有点相连，费用cost[s][u]=0（这是显然的），容量cap[s][u]=1；V中所有点与超级汇t相连，

费用cost[v][t]=0（这是显然的），容量cap[t][v]=1。

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
#include <cmath>
using namespace std;
const int INF=1<<29;
const int maxn=210;
struct node
{
    int x,y;
}p[maxn],h[maxn];
int c[maxn][maxn],f[maxn][maxn],w[maxn][maxn];
int n,m,pn,hn,s,t,ans;
bool inq[maxn];  //是否在队列中
char str[maxn][maxn];
int pre[maxn],dis[maxn];//上一条弧，Bellman_ford；

void spfa()
{
    int i,v;
    queue<int>q;
    for (i = 0; i < maxn; i++)
    {
        dis[i] = INF;
        pre[i] = -1;
        inq[i] = false;
    }
    q.push(s);
    inq[s] = true;
    dis[s] = 0;
    while (!q.empty())
    {
        int u = q.front();
        q.pop();
        inq[u] = false;
        for (v = 0; v <= t; v++)
        {
            if (c[u][v]&& dis[v] > dis[u] + w[u][v])
            {
                dis[v] = dis[u] + w[u][v];
                pre[v] = u;
                if (!inq[v])
                {
                    q.push(v);
                    inq[v] = true;
                }
            }
        }
    }
}
void mcmf()
{
    while (1)
    {
        spfa();
        if (pre[t] == -1) break;
        int x = t,minf = INF;
        while (pre[x] != -1)
        {
            minf = min(minf,c[pre[x]][x]);
            x = pre[x];
        }
        x = t;
        while (pre[x] != -1)
        {
            c[pre[x]][x]-=minf;
            c[x][pre[x]]+=minf;
            ans+=minf*w[pre[x]][x];
            x = pre[x];
        }
    }
}

int main()
{
    int i,j,pn,hn;
    while(cin>>n>>m && n || m)
    {
        pn = hn = 0;
        memset(c,0,sizeof(c));
        memset(f,0,sizeof(f));
        memset(w,0,sizeof(w));
        for(i = 0; i < n; i++)
        {
            scanf("%s",str[i]);
            for(j = 0; j < m; j++)
            {
                if(str[i][j] == 'H')
                {
                    h[++hn].x = i;
                    h[hn].y = j;
                }
                else if(str[i][j] == 'm')
                {
                    p[++pn].x = i;
                    p[pn].y = j;
                }
            }
        }
        s = 0;
        t = pn + hn + 1;
        for (i = 1; i <= pn; i++)
        c[s][i] = 1;
        for (i = 1; i <= hn; i++)
        c[i + pn][t] = 1;
        for (i = 1; i <= pn; i++)
        {
            for (j = 1; j <= hn; j++)
            {
                c[i][j + pn] = 1;
                w[i][j + pn] = abs(p[i].x - h[j].x) + abs(p[i].y - h[j].y);
                w[j + pn][i] = -w[i][j + pn];
            }

        }
        ans = 0;
        mcmf();
        printf("%d
",ans);
    }
    return 0;
}