pku 2195 Going Home 最小费最大流问题

http://poj.org/problem?id=2195

题意是：有相同数量的人与房子，每一时刻人都可以花费1$的钱走一步，问让每个人到达一个屋子的最少需要的费用。

建立源点与汇点，求有源点到汇点的最小费用最大流；改了一下不需要f[][]的模板。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#include <cstdlib>
#define maxn 207
using namespace std;

const int inf = 999999999;

struct node
{
    int x,y;
}p[maxn],h[maxn];
int c[maxn][maxn],f[maxn][maxn],w[maxn][maxn];
int pre[maxn],dis[maxn];
int n,m,pn,hn,s,t;
char str[maxn][maxn];
bool inq[maxn];
int ans;

int Abs(int x)
{
    return x > 0 ? x : -x;
}
void spfa()
{
    int v;
    queue<int>q;
    for (int 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()
{
    //freopen("in.txt","r",stdin);
    int i,j,pn,hn;
    while (~scanf("%d%d",&n,&m))
    {
        if (!n && !m) break;
        pn = hn = 0;
        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;
                }
            }
        }
        memset(c,0,sizeof(c));
        memset(f,0,sizeof(f));
        memset(w,0,sizeof(w));
        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\n",ans);
    }
    return 0;
}