poj 2915 Going Home (最小费用流)

版权声明：本文为博主原创文章，未经博主允许不得转载。 https://blog.csdn.net/qq_40993793/article/details/81484280

传送门：http://poj.org/problem?id=2195

  有相同人数的人和房子，每走一步的花费为1，问每个人都到一个互不相同的房子里最少的花费为多少。

  增加一个超级源点S和超级汇点T，S连向人(容量1，cost 0)；房子连向T(容量1，cost 0)；每个人都连向所有房子(容量1，cost人与房子的距离|xi-yi|+|xj-yj|)。

  建好图后就成了最小费用流模板题，最小费用流跑一遍就可以了。

  这题给的数据范围有点问题，尽量往大的开吧。。

AC代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<cstdlib>
#include<utility>
#include<algorithm>
#include<utility>
#include<queue>
#include<vector>
#include<set>
#include<stack>
#include<cmath>
#include<map>
#include<ctime>
#include<functional>
#define P pair<int,int>
#define ll long long
#define ull unsigned long long
#define lson id*2,l,mid
#define rson id*2+1,mid+1,r
#define ls id*2
#define rs id*2+1
#define Mod(a,b) a<b?a:a%b+b 
using namespace std;

const ll M = 1e9 + 7;
const ll INF = 1e9;
const int N = 10010;
const double e = 10e-6;

int n, m;

struct edge { int to, cap, cost, rev; };
vector<edge> G[N];
int V;
int dist[N], prevv[N], preve[N], h[N];
void addEdge(int from, int to, int cap, int cost)
{
	G[from].push_back(edge{ to,cap,cost,(int)G[to].size() });
	G[to].push_back(edge{ from,0,-cost,(int)G[from].size() - 1 });
}
int minCost_D(int s, int t, int f)
{
	int res = 0;
	memset(h, 0, sizeof(h));
	while (f > 0) {
		priority_queue<P, vector<P>, greater<P> > que;
		fill(dist, dist + V, INF);
		dist[s] = 0;
		que.push(P(0, s));
		while (!que.empty()) {
			P p = que.top(); que.pop();
			int v = p.second;
			if (dist[v] < p.first) continue;
			for (int i = 0; i < G[v].size(); i++) {
				edge &e = G[v][i];
				if (e.cap > 0 && dist[e.to] > dist[v] + e.cost + h[v] - h[e.to]) {
					dist[e.to] = dist[v] + e.cost + h[v] - h[e.to];
					prevv[e.to] = v; preve[e.to] = i;
					que.push(P(dist[e.to], e.to));
				}
			}
		}
		if (dist[t] == INF) return -1;
		for (int v = 0; v < V; v++)
			h[v] += dist[v];
		int d = f;
		for (int v = t; v != s; v = prevv[v])
			d = min(d, G[prevv[v]][preve[v]].cap);
		f -= d;
		res += d*h[t];
		for (int v = t; v != s; v = prevv[v]) {
			edge &e = G[prevv[v]][preve[v]];
			e.cap -= d;
			G[v][e.rev].cap += d;
		}
	}
	return res;
}

P man[5010], house[5010];
int cnt1, cnt2;

int main() {
	int u, v, S, T;
	char c;
	while (scanf("%d%d", &n, &m)) {
		if (n == 0 && m == 0)
			break;	
		getchar(); cnt1 = 0; cnt2 = 0;
		for (int i = 1; i <= n; i++) {
			for (int j = 1; j <= m; j++) {
				c = getchar();
				if (c == 'm')
					man[cnt1++] = P(i, j);
				else if (c == 'H')
					house[cnt2++] = P(i, j);
			}
			getchar();
		}
		S = 0; T = 2 * cnt1 + 1;
		V = 2 * cnt1 + 2;
		for (int i = 0; i < cnt1; i++)
			addEdge(S, i + 1, 1, 0);
		for (int i = 0; i < cnt1; i++)
			addEdge(i + cnt1 + 1, T, 1, 0);
		for (int i = 0; i < cnt1; i++)
			for (int j = 0; j < cnt1; j++) {
				int x = abs(man[i].first - house[j].first) + abs(man[i].second - house[j].second);
				addEdge(i + 1, j + cnt1 + 1, 1, x);
			}

		printf("%d\n", minCost_D(S, T, cnt1));

		for (int i = 0; i <= 2 * cnt1 + 1; i++)
			G[i].clear();
	}

	return 0;
}