poj2195 Going Home，最小费用最大流

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;}