poj 2135 Farm Tour (最小费用流)

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传送门：http://poj.org/problem?id=2135

题意：

  n个顶点，m条无向边，从顶点0到顶点n找出2条路使总距离最短，这2条路中没有重复使用的边。

思路：

  建图，每条边的流量都为1，那么流量为2的最小费用流就是答案。

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 int maxn = 100010;
const ll M = 1e9 + 7;
const ll INF = 1e9;
const int N = 1010;
const double e = 10e-6;
const int dx[4] = { 0,0,1,-1 }, dy[4] = { 1,-1,0,0 };

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 });
}
void init()
{
	memset(prevv, 0, sizeof(prevv));
	memset(preve, 0, sizeof(preve));
	for (int i = 0; i < N; i++)
		G[i].clear();
}
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;
}

int n, m, a, b, x;

int main() {
	int S = 0, T = 1;
	while (~scanf("%d%d", &n, &m)) {
		init(); V = n;
		for (int i = 1; i <= m; i++) {
			scanf("%d%d%d", &a, &b, &x);
			addEdge(a - 1, b - 1, 1, x);
			addEdge(b - 1, a - 1, 1, x);
		}
		printf("%d\n", minCost_D(0, n - 1, 2));
	}
	return 0;
}