【POJ - 2135 】【最小费用最大流】【裸】【1->n->1不重复路的最小值】

http://poj.org/problem?id=2135

【题意】

求1->n->1不重复路的最短路

【题解】最小费用最大流的模板题

               最小费用最大流-->多次s'p'fa-->满足最大情况下，用spfa找可行流的最短路径

               建图：源点s:0,汇点t:n+1

               1.add(s,1,2(流量),0（费用）)

               2.add(n,t,2,0)

               3.add(u,v,1,0)

                  add(v,u,1,0)

【代码】

#include<iostream>
#include <algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
#include<vector>
using namespace std;
typedef long long ll;
const int INF = 100000000;
const int N = 1005;
struct edge
{
	int from, to, cap, flow, cost;
	edge(int u, int v, int c, int f, int co) :from(u), to(v), cap(c), flow(f), cost(co) {}
};
struct MCMF
{
	int n, m;
	vector<edge>edges;
	vector<int>g[N];
	int inq[N];
	int d[N];
	int p[N];
	int a[N];
	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 spfa(int s, int t, int &flow, int &cost) {
		for (int i = 0; i <= n; i++)d[i] = INF;
		memset(inq, 0, sizeof(inq));
		d[s] = 0; inq[s] = 1; p[s] = 0; a[s] = INF;
		queue<int>q;
		q.push(s);
		while (!q.empty()) {
			int u = q.front(); q.pop();
			inq[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 (!inq[e.to]) { q.push(e.to); inq[e.to] = 1; }
				}
			}
		}
		if (d[t] == INF)return 0;
		flow += a[t];
		cost += d[t] * a[t];
		for (int u = t; u != s; u = edges[p[u]].from) {
			edges[p[u]].flow += a[t];
			edges[p[u] ^ 1].flow -= a[t];
		}
		return 1;
	}
	int MincoatMaxflow(int s, int t, int& cost) {
		int flow = 0; cost = 0;
		while (spfa(s, t, flow, cost));
		return flow;
	}
};
int main()
{
	int m, n;
	MCMF mcmf;
	scanf("%d%d", &n, &m);
	mcmf.init(n + 1);
	while (m--) {
		int u, v, w;
		scanf("%d%d%d", &u, &v, &w);
		mcmf.addedge(u, v, 1, w);
		mcmf.addedge(v, u, 1, w);
	}
	mcmf.addedge(0, 1, 2, 0);
	mcmf.addedge(n, n + 1, 2, 0);
	n++;
	int cost = 0;
	int ans = mcmf.MincoatMaxflow(0, n, cost);
	cout << cost << endl;
}