学步园

题意：

给定一个N*M的地图，地图上有若干个man和house，且man与house的数量一致。man每移动一格需花费$1（即单位费用=单位距离），一间house只能入住一个man。现在要求所有的man都入住house，求最小费用。

分析：

二分图的最大匹配

我采用的是最小费用最大流算法，重点在建图。

Code:

#include <cmath>
#include <cstdio>
#include <cstring>
#include <queue>
#include <vector>
#include <algorithm>
using namespace std;

const int maxn = 200 + 10;
const int INF = 1000000000;
typedef long long LL;
int msum, hsum;
struct xy {
    int x, y;
    xy(int x = 0, int y = 0): x(x), y(y) {}
};
xy M[maxn], H[maxn];

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

struct MCMF {
    int n, m, s, t;
    vector<Edge> edges;
    vector<int> G[maxn];
    int vis[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    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& cost) {
        for (int i = 0; i <= n; i++) d[i] = INF;
        memset(vis, 0, sizeof(vis));
        d[s] = 0; vis[s] = 1; p[s] = 0; a[s] = INF;

        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];
                    a[e.to] = min(a[u], e.cap - e.flow);
                    if (!vis[e.to]) { Q.push(e.to); vis[e.to] = 1;}
                }
            }
        }
        if (d[t] == INF) return false;
        cost += d[t] * a[t];
        int u = t;
        while (u != s) {
            edges[p[u]].flow += a[t];
            edges[p[u] ^ 1].flow -= a[t];
            u = edges[p[u]].from;
        }
        return true;
    }

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

MCMF g;
int n, m, s, t;
int dis(xy& a, xy& b) {
    return abs(a.x - b.x) + abs(a.y - b.y);
}
void make_graph() {
    int i, j, cost;
    char str[maxn];
    msum = hsum = 0;
    for (i = 0; i < n; i++) {
        scanf("%s", str);
        for (j = 0; j < m; j++)
            if (str[j] == 'm') {
                M[msum++] = xy(i, j);
            } else if (str[j] == 'H') {
                H[hsum++] = xy(i, j);
            }
    }
    s = msum + hsum + 1, t = msum + hsum + 2;
    g.init(t);
    for (i = 0; i < msum; i++)
        for (j = 0; j < hsum; j++) {
            cost = dis(M[i], H[j]);
            g.AddEdge(i, j + msum, 1, cost);
        }

    for (i = 0; i < msum; i++) g.AddEdge(s, i , 1, 0);
    for (i = 0; i < hsum; i++) g.AddEdge(i + msum, t, 1, 0);

};
int main() {
    while (scanf("%d%d", &n, &m)) {
        if (n == 0 && m == 0) break;
        make_graph();
        int answer = g.Mincost(s, t);
        printf("%d\n", answer);
    }
    return 0;
}