POJ - 2195 Going Home （MCMF）

题目大意:在一张地图上，有n个人和n间房间，现在要求将这n个人移动到这n间房子里面，移动一次的代价是1，每间房子最终只能属于1个人，问最少的移动代价是多少

解题思路：
超级源点–人，容量为1，费用为0
人—房子，容量为1，费用为移动距离
房子—超级汇点，容量为1，费用为0

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <vector>
using namespace std;
#define N 1010
#define INF 0x3f3f3f3f

struct Edge{
    int from, to, cap, flow ,cost;
    Edge() {}
    Edge(int from, int to, int cap, int flow, int cost):from(from), to(to), cap(cap), flow(flow), cost(cost) {}
};

struct MCMF{
    int n, m, source, sink;
    vector<Edge> edges;
    vector<int> G[N];
    int d[N], f[N], p[N];
    bool vis[N];

    void init(int n) {
        this->n = n;
        for (int i = 0; i <= n; i++)
            G[i].clear();
        edges.clear();
    }

    void AddEdge(int from, int to, int cap, int cost) {
        edges.push_back(Edge(from, to, cap, 0, cost));
        edges.push_back(Edge(to, from, 0, 0, -cost));
        m = edges.size();
        G[from].push_back(m - 2);
        G[to].push_back(m - 1);
    }

    bool BellmanFord(int s, int t, int &flow, int &cost) {
        for (int i = 0; i <= n; i++)
            d[i] = INF;
        memset(vis, 0, sizeof(vis));
        vis[s] = 1; d[s] = 0; f[s] = INF; p[s] = 0;
        queue<int> Q;
        Q.push(s);

        while (!Q.empty()) {
            int u = Q.front();
            Q.pop();
            vis[u] = 0;

            for (int i = 0; i < G[u].size(); i++) {
                Edge &e = edges[G[u][i]];
                if (e.cap > e.flow && d[e.to] > d[u] + e.cost) {
                    d[e.to] = d[u] + e.cost;
                    p[e.to] = G[u][i];
                    f[e.to] = min(f[u], e.cap - e.flow);
                    if (!vis[e.to]) {
                        vis[e.to] = true;
                        Q.push(e.to);
                    }
                }
            }
        }

        if (d[t] == INF)
            return false;

        flow += f[t];
        cost += d[t];

        int u = t;
        while (u != s) {
            edges[p[u]].flow += f[t];
            edges[p[u] ^ 1].flow -= f[t];
            u = edges[p[u]].from;
        }
        return true;
    }

    int Mincost(int s, int t) {
        int flow = 0, cost = 0;
        while (BellmanFord(s, t, flow, cost));
        return cost;
    }
};

//超级源点 -- 人 容量为1，费用为0
//人---房子，容量为1，费用为移动距离
//房子--超级汇点,容量为1，费用为0

MCMF mcmf;
#define M 110
#define ABS(x) ((x) > 0 ? x : (-(x)))
struct Man{
    int x, y;
}man[M];

struct House{
    int x, y;
}house[M];

char  map[M][M];
int n, m;

void solve() {
    for (int i = 0; i < n; i++)
        scanf("%s", map[i]);

    int cnt_m = 0, cnt_h = 0;
    for (int i = 0; i < n; i++)
        for (int j = 0; j < m; j++) {
            if (map[i][j] == ‘m‘) {
                man[cnt_m].x = i;
                man[cnt_m++].y = j;
            }
            else if(map[i][j] == ‘H‘) {
                house[cnt_h].x = i;
                house[cnt_h++].y = j;
            }
        }
    if (cnt_m + cnt_h == 0) {
        printf("0\n");
        return ;
    }

    int s = cnt_m + cnt_h, t = cnt_m + cnt_h + 1;
    mcmf.init(t);

    for (int i = 0; i < cnt_m; i++)  
        mcmf.AddEdge(s, i, 1, 0);

    for (int i = 0; i < cnt_h; i++)
        mcmf.AddEdge(cnt_m + i, t, 1, 0);

    for (int i = 0; i < cnt_m; i++) {
        for (int j = 0; j < cnt_h; j++) {
            int x = ABS(man[i].x - house[j].x);
            int y = ABS(man[i].y - house[j].y);

            mcmf.AddEdge(i, cnt_m + j, 1, x + y);
        }
    }
    printf("%d\n", mcmf.Mincost(s, t));
}

int main() {
    while (scanf("%d%d", &n, &m) != EOF && n + m) {
        solve();
    }
    return 0;
}