POJ 2195 Going Home 题解（最小费用最大流）

依然是很常规的最小费用流，新建源点s，从s到每个人建立cap=1,cost=0的边；新建汇点t，从每个房子建立cap=1,cost=0的边到t；从每个人建立相应费用，cap=1的边到每个房子，然后跑F=人数的最小费用流即可.

#include<iostream>
#include<vector>
#include<functional>
#include<queue>
#include<cstring>
#include<cstdio>
#include<algorithm>
using namespace std;
const int MAX_N = 100 + 16;
const int MAX = MAX_N * MAX_N * 2;
const int INF = 0x3f3f3f3f;
typedef pair<int, int> P;
struct node{
	int to;
	int cap;
	int cost;
	int rev;
	node(int to, int cap, int cost, int rev) :to(to), cap(cap), cost(cost), rev(rev){}
};
int n;
vector<node> G[MAX];
int h[MAX];
int dis[MAX];
int prevv[MAX];
int preve[MAX];
void addNode(int from, int to, int cap, int cost){
	G[from].push_back(node(to, cap, cost, G[to].size()));
	G[to].push_back(node(from, 0, - cost, G[from].size() - 1));
}
int min_cost_flow(int s, int t, int f){
	int ret = 0;
	memset(h, 0, sizeof(h));
	while(f > 0){
		priority_queue<P, vector<P>, greater<P> > q;
		fill(dis, dis + MAX, INF);
		dis[s] = 0;
		q.push(P(0, s));
		while(!q.empty()){
			P p = q.top(); q.pop();
			int v = p.second;
			if(dis[v] < p.first) continue;
			for(int i = 0; i < G[v].size(); ++i){
				node &e = G[v][i];
				if(e.cap > 0 && dis[e.to] > dis[v] + e.cost + h[v] - h[e.to]){
					dis[e.to] = dis[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v;
					preve[e.to] = i;
					q.push(P(dis[e.to], e.to));
				}
			}
		}
		if(dis[t] == INF)	return -1;

		for(int v = 0; v < n; ++v)	h[v] += dis[v];
		int d = f;
		for(int v = t; v != s; v = prevv[v])	d = min(d, G[prevv[v]][preve[v]].cap);
		f -= d;
		ret += d * h[t];
		for(int v = t; v != s; v = prevv[v]){
			node &e = G[prevv[v]][preve[v]];
			e.cap -= d;
			G[e.to][e.rev].cap += d;
		}
	}
	return ret;
}
int calculateDis(P a, P b){
	return abs(a.first - b.first) + abs(a.second - b.second);
}
int main(){
	int N, M;
	while(~scanf("%d%d", &N, &M), N && M){
		vector<P> man;
		vector<P> house;
		for(int i = 0; i < N; ++i){
			getchar();
			for(int j = 0; j < M; ++j){
				switch(getchar()){
					case 'm' : man.push_back(P(i, j)); break;
					case 'H' : house.push_back(P(i, j)); break;
				}
			}
		}
		const int size = man.size();
		const int s = 0, t = 1 + size * 2;
		n = t + 1;
		memset(G, 0, sizeof(G));
		for(int i = 0; i < size; ++i){
			addNode(s, i + 1, 1, 0);
			addNode(i + 1 + size, t, 1, 0);
			for(int j = 0; j < size; ++j)
				addNode(i + 1, size + j + 1, 1, calculateDis(man[i],house[j]));
		}
		printf("%d\n", min_cost_flow(s, t, size));
	}
	return 0;
}