Hdu1533 Going Home 最小费用最大流

题意：用H来代表人类，m来代表房子，问每个人类到每个房子的总的 最短距离 。
思路：这里注意价值是H到M要走的距离，而容量的话，用1来表示就好了，因为H可以到达任何一个M.
剩下直接套模板就行，注意字符判断时不要带有空格。
最小费用流总结：1_尽量不要要用disk或者floyed，容易超时。2_输入时取两条边，价值取反。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#define mem(a,b) memset(a,b,sizeof(a))
using namespace std;
const int maxn = 100010;
bool vis[maxn];
int n, m, s, t, dis[maxn], pre[maxn], last[maxn], flow[maxn], maxflow, mincost;
//dis最小花费;pre每个点的前驱；last每个点的所连的前一条边；flow源点到此处的流量
struct Edge {
    int to, next, flow, dis; //flow流量 dis花费
} edge[maxn];
struct node {
    int x;
    int y;
    int w;
} H[maxn];
struct Node {
    int x;
    int y;
    int w;
} M[maxn];
int head[maxn], num_edge = 1;
queue <int> q;
void add_edge(int from, int to, int flow, int dis) {
    edge[++num_edge].next = head[from];
    edge[num_edge].to = to;
    edge[num_edge].flow = flow;
    edge[num_edge].dis = dis;
    head[from] = num_edge;
}
bool spfa(int s, int t) {
    mem(dis, 0x7f);
    mem(flow, 0x7f);
    mem(vis, 0);
    q.push(s);
    vis[s] = 1;
    dis[s] = 0;
    pre[t] = -1;
    while (!q.empty()) {
        int now = q.front();
        q.pop();
        vis[now] = 0;
        for (int i = head[now]; i != -1; i = edge[i].next) {
            if (edge[i].flow > 0 && dis[edge[i].to] > dis[now] + edge[i].dis) { //正边
                dis[edge[i].to] = dis[now] + edge[i].dis;
                pre[edge[i].to] = now;
                last[edge[i].to] = i;
                flow[edge[i].to] = min(flow[now], edge[i].flow); //
                if (!vis[edge[i].to]) {
                    vis[edge[i].to] = 1;
                    q.push(edge[i].to);
                }
            }
        }
    }
    return pre[t] != -1;
}

void MaxFlow_MinCost() { ///最大流的最小费用
    while (spfa(s, t)) {
        int now = t;
        maxflow += flow[t];
        mincost += dis[t];
        while (now != s) {
            //从源点一直回溯到汇点
            edge[last[now]].flow -= flow[t]; //flow和dis容易搞混
            edge[last[now] ^ 1].flow += flow[t];
            now = pre[now];
        }
    }
}

int main() {
    while(cin >> n >> m,n||m){
    num_edge = 1;
    mincost = 0;
    mem(head, -1);
    num_edge = -1; //初始化
    s = 0, t = 206;
    char u ;
    int cnt_1 = 100, cnt_2 = 0 ;
    for(int i = 1; i <= n; i++)
        for(int j = 1; j <= m; j++) {
            cin >> u;
            if(u == 'm') {
                ++cnt_1;
                M[cnt_1].x = i;
                M[cnt_1].y = j;
            } else if(u == 'H') {
                ++cnt_2;
                H[cnt_2].x = i;
                H[cnt_2].y = j;
            }
        }
    for(int i = 1; i <= cnt_2; i++)
        add_edge(s, i, 1, 0), add_edge(i, s, 0, 0);
    for(int i = 101; i <= cnt_1; i++)
        add_edge(i, t, 1, 0), add_edge(t, i, 0, 0);
    for(int i = 1; i <= cnt_2; i++)
        for(int j = 101; j <= cnt_1; j++) {
            int w = abs(M[j].x - H[i].x) + abs(M[j].y - H[i].y);
            add_edge(i, j, 1, w);
            add_edge(j, i, 0, -w);
        }
    MaxFlow_MinCost();
    cout << mincost << endl;
    }
    return 0;
}