poj2195 Going Home

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;

}