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
4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2Sample Output
6思路：
这道题一开始想用两次dijkstra解决，但其实是不行的因为有可能找到一条最短路后，第二次就没有路径可走了，但第一次走的不是最短路，之后可能会有路径，这一题是要求到再回去所以不行。如下图：
看完题解后才知道这是一道最小费用最大流的题目，最小费用最大流，先是保证最大流量在找最小的费用，用spfa找最短路。这道题目中因为每条路只能走一遍，所以所有的边的流量都设为1，又因为是找两条路，所以找到两条增广路后，及流量为2时就退出搜索。
以下来自：此博客
求解步骤
（1）找到一条从源点到达汇点的“距离最短”的路径，“距离”使用该路径上的边的单位费用之和来衡量。
（2）然后找出这条路径上的边的容量的最小值f，则当前最大流max_flow扩充f，同时当前最小费用min_cost扩充 f*min_dist(s,t)。
（3）将这条路径上的每条正向边的容量都减少f，每条反向边的容量都增加f。
（4）重复（1）--（3）直到无法找到从源点到达汇点的路径。
ac代码：
#include<stdio.h>
#include<string.h>
#include<queue>
#include<set>
#include<iostream>
#include<map>
#include<stack>
#include<cmath>
#include<algorithm>
#define ll long long
#define mod 1000000007
#define eps 1e-8
using namespace std;
const int MAXN = 10000;
const int MAXM = 100000;
const int INF = 0x3f3f3f3f;
struct Edge {
int to,next,cap,flow,cost;
} edge[MAXM];
int head[MAXN],tol;
int pre[MAXN],dis[MAXN];
bool vis[MAXN];
int N;//节点总个数，节点编号从 0 ～ N-1
void init(int n) {
N = n;
tol = 0;
memset(head,-1,sizeof(head));
}
void addedge(int u,int v,int cap,int cost) {
edge[tol].to = v;
edge[tol].cap = cap;
edge[tol].cost = cost;
edge[tol].flow = 0;
edge[tol].next = head[u];
head[u] = tol++;
edge[tol].to = u;
edge[tol].cap = 0;
edge[tol].cost = -cost;
edge[tol].flow = 0;
edge[tol].next = head[v];
head[v] = tol++;
}
bool spfa(int s,int t) {
queue<int>q;
for(int i = 0; i < N+1; i++) {
dis[i] = INF;
vis[i] = false;
pre[i] = -1;
}
dis[s] = 0;
vis[s] = true;
q.push(s);
while(!q.empty()) {
int u = q.front();
q.pop();
vis[u] = false;
for(int i = head[u]; i != -1; i = edge[i].next) {
int v = edge[i].to;
if(edge[i].cap > edge[i].flow && dis[v] > dis[u] + edge[i].cost ) {
dis[v] = dis[u] + edge[i].cost;
pre[v] = i;
if(!vis[v]) {
vis[v] = true;
q.push(v);
}
}
}
}
if(pre[t] == -1)return false;
else return true;
}
//返回的是最大流，cost 存的是最小费用
int minCostMaxflow(int s,int t,int &cost) {
int flow = 0;
cost = 0;
while(spfa(s,t)) {
int Min = INF;
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
if(Min > edge[i].cap - edge[i].flow)
Min = edge[i].cap - edge[i].flow;
}
for(int i = pre[t]; i != -1; i = pre[edge[i^1].to]) {
edge[i].flow += Min;
edge[i^1].flow -= Min;
cost += edge[i].cost * Min;
}
flow += Min;
if(flow==2)
return flow;
}
return flow;
}
int main()
{
int n,m;
scanf("%d%d",&n,&m);
init(n);
for(int i=0;i<m;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,1,w);
addedge(v,u,1,w);
}
int c;
minCostMaxflow(1,n,c);
cout<<c<<endl;
return 0;
}