poj2135 Farm Tour 最小费用流

原创于 2019-09-16 21:22:35 发布 · 150 阅读

CC 4.0 BY-SA版权

本文介绍了一种求解从起点到终点的两条无公共边最短路径的问题，通过转化为求最小费用流的方法来解决。利用SPFA算法进行最短路径搜索，并通过多次迭代找到总长度最短的两条路径。

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 21810		Accepted: 8364

Description

When FJ's friends visit him on the farm, he likes to show them around. His farm comprises N (1 <= N <= 1000) fields numbered 1..N, the first of which contains his house and the Nth of which contains the big barn. A total M (1 <= M <= 10000) paths that connect the fields in various ways. Each path connects two different fields and has a nonzero length smaller than 35,000.

To show off his farm in the best way, he walks a tour that starts at his house, potentially travels through some fields, and ends at the barn. Later, he returns (potentially through some fields) back to his house again.

He wants his tour to be as short as possible, however he doesn't want to walk on any given path more than once. Calculate the shortest tour possible. FJ is sure that some tour exists for any given farm.

Input

* Line 1: Two space-separated integers: N and M.

* Lines 2..M+1: Three space-separated integers that define a path: The starting field, the end field, and the path's length.

Output

A single line containing the length of the shortest tour.

Sample Input

Sample Output

Source

USACO 2003 February Green

/*将问题转化为求从1号顶点到N号顶点的两条没有公共边的路径，转化后即求流量为2的最小费用流了。*/
#include <iostream>
#include <algorithm>
#include <cstring>
#include <vector>
#include <queue>
#include <cstdio>
using namespace std;
typedef long long ll;
const int N=1005;
const int INF=0x3f3f3f3f;
int n,m,s,t;
struct edge{
	ll to,cap,cost,rev;
};
vector<edge>vec[N];
ll dis[N],vis[N];
int preV[N],preE[N];
void add_edge(int from,int to,ll cap,ll cost){
	vec[from].push_back(edge{to,cap,cost,vec[to].size()});
	vec[to].push_back(edge{from,0,-cost,vec[from].size()-1});
}
bool spfa(int s,int t){
	memset(preV,-1,sizeof(preV));
	memset(dis,INF,sizeof(dis));
	dis[s]=0;
	vis[s]=true;
	queue<int>que;
	que.push(s);
	while(!que.empty()){
		int now=que.front();
		que.pop(); vis[now]=false;
		for(int i=0;i<vec[now].size();i++){
			edge &e=vec[now][i];
			if(dis[now]+e.cost<dis[e.to]&&e.cap>0){
				dis[e.to]=dis[now]+e.cost;
				preV[e.to]=now;
				preE[e.to]=i;
				if(!vis[e.to]){//必须在这里判断标记，不然超时。
					que.push(e.to);
					vis[e.to]=true;	
				}
				
			}
		}
	}
	if(preV[t]==-1) return false;
	return true;
}
ll min_cost_flow(int s,int t,ll f){
	ll cost=0;
	while(spfa(s,t)&&f>0){
		ll d=INF;
		for(int v=t;v!=s;v=preV[v]){
			d=min(d,vec[preV[v]][preE[v]].cap);
		}
		f-=d;
		cost+=dis[t]*d;
		for(int v=t;v!=s;v=preV[v]){
			edge &e=vec[preV[v]][preE[v]];
			e.cap-=d;
			vec[v][e.rev].cap+=d;
		}
	}
	return cost;
}
int main(){
	int a,b,c;
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++){
		scanf("%d %d %d",&a,&b,&c);
		add_edge(a,b,1,c);
		add_edge(b,a,1,c);
	}
	printf("%d\n",min_cost_flow(1,n,2));
	return 0;
}