Farm Tour POJ - 2135 

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 2

Sample Output

6

題意很簡單，說思路；

 

設源點為s，匯點為t；

然後建圖的時候，s->1   容量為2，費用為0；

n->t，容量為2，費用為0；

這樣就能從s出發，到t，然後再走不重複的路徑的再回到s；

由於s->1,和n->t這兩條路徑費用為0，所以不影響最後結果。

這樣直接跑一邊最大費用最小流的模板就行了。

詳見程式碼：

 

  1 #include <iostream>
  2 #include <cstdio>
  3 #include <cstring>
  4
 
 #include <algorithm>
  5 #include <vector>
  6 #include <queue>
  7 using namespace std;
  8 typedef long long ll;
  9 const int MAXX=40010;
 10 const int INF=0x3f3f3f3f;
 11 
 12 struct node
 13 {
 14     int to;
 15     int next;
 16     int cap;
 17     int flow;
 18     int cost;
 19 }edge[MAXX];
 20 
 21 int head[MAXX],tol;
 22 int pre[MAXX],dis[MAXX];
 23 bool vis[MAXX];
 24 int N;
 25 
 26 void init(int n)
 27 {
 28     N=n;
 29     tol=0;
 30     memset(head,-1,sizeof(head));
 31 }
 32 
 33 void addedge(int u,int v,int cap,int cost)
 34 {
 35     edge[tol].to=v;
 36     edge[tol].cap=cap;
 37     edge[tol].cost=cost;
 38     edge[tol].flow=0;
 39     edge[tol].next=head[u];
 40     head[u]=tol++;
 41     edge[tol].to=u;
 42     edge[tol].cap=0;
 43     edge[tol].cost=-cost;
 44     edge[tol].flow=0;
 45     edge[tol].next=head[v];
 46     head[v]=tol++;
 47 }
 48 
 49 bool SPFA(int s,int t)
 50 {
 51     queue<int> q;
 52     for(int i=0;i<N;i++)
 53     {
 54         dis[i]=INF;
 55         vis[i]=0;
 56         pre[i]=-1;
 57     }
 58     dis[s]=0;
 59     vis[s]=1;
 60     q.push(s);
 61     while(!q.empty())
 62     {
 63         int u=q.front();
 64         q.pop();
 65         vis[u]=0;
 66         for(int i=head[u];i!=-1;i=edge[i].next)
 67         {
 68             int v=edge[i].to;
 69             if(edge[i].cap>edge[i].flow&&dis[v]>dis[u]+edge[i].cost)
 70             {
 71                 dis[v]=dis[u]+edge[i].cost;
 72                 pre[v]=i;
 73                 if(!vis[v])
 74                 {
 75                     vis[v]=1;
 76                     q.push(v);
 77                 }
 78             }
 79         }
 80     }
 81     if(pre[t]==-1)return 0;
 82     return 1;
 83 }
 84 
 85 int minCostMaxFlow(int s,int t)
 86 {
 87     int cost=0;
 88     while(SPFA(s,t))
 89     {
 90         int minn=INF;
 91         for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
 92         {
 93             if(minn>edge[i].cap-edge[i].flow)
 94                 minn=edge[i].cap-edge[i].flow;
 95         }
 96         for(int i=pre[t];i!=-1;i=pre[edge[i^1].to])
 97         {
 98             edge[i].flow+=minn;
 99             edge[i^1].flow-=minn;
100             cost+=edge[i].cost*minn;
101         }
102     }
103     return cost;
104 }
105 int main()
106 {
107 
108     int n,m;
109     while(~scanf("%d%d",&n,&m))
110     {
111         int u,v,g;
112         init(n+2);
113         for(int i=0;i<m;i++)
114         {
115 
116             scanf("%d%d%d",&u,&v,&g);
117             addedge(u,v,1,g);
118             addedge(v,u,1,g);
119         }
120         addedge(0,1,2,0);
121         addedge(n,n+1,2,0);
122         printf("%d\n",minCostMaxFlow(0,n+1));
123     }
124     return 0;
125 }