POJ 2135 最小费用最大流 入门题

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 19207		Accepted: 7441

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

4 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

解析 题目是无向图 求两条最短路径的长度 ,两条路径不允许有相交边。其实就是求一条最短路，把边删掉，再找一条。其实可以转化为网络流来写。

无向图把我们建边要建两次ab，ba；ba，ab(模板可以有平行边)   把每个边的容量设为1 花费为边的长度。然后找两次增广路退出来的花费就是答案。

也可以建立超源点汇点 0,n+1   建立两条边  0—1 容量为2 花费为0   n—n+1 容量为2 花费为0   跑一边最小费用就是答案了

锦囊：

以下板子求的是最小费用最大流 如果要固定流量k 可以在增广的时候检查一下 在flow+a>=k 的时候只增广k-flow单位的流量，然后终止程序。同样有费用限制也一样。

 #include<iostream>
 #include<stdio.h>
 #include<string.h>
 #include<stdlib.h>
 #include<vector>
 #include<queue>
 using namespace std;
 const int maxn=1e3+,mod=1e9+,inf=0x3f3f3f3f;
 struct edge
 {
     int to,next,cap,flow,cost;
 } edge[maxn*maxn];
 int head[maxn],tol;
 int pre[maxn],dis[maxn];
 bool vis[maxn];
 int N;
 void init(int n)
 {
     N=n,tol=;
     memset(head,-,sizeof(head));
 }
 void addedge(int u,int v,int cap,int cost)
 {
     edge[tol].to=v;
     edge[tol].cap=cap;
     edge[tol].flow=;
     edge[tol].cost=cost;
     edge[tol].next=head[u];
     head[u]=tol++;
     edge[tol].to=u;
     edge[tol].cap=;
     edge[tol].flow=;
     edge[tol].cost=-cost;
     edge[tol].next=head[v];
     head[v]=tol++;
 }
 bool spfa(int s,int t)
 {
     queue<int> q;
     for(int i=; i<=N; i++)
     {
         dis[i]=inf;
         vis[i]=false;
         pre[i]=-;
     }
     dis[s]=;
     vis[s]=true;
     q.push(s);
     while(!q.empty())
     {
         int u=q.front();
         q.pop();
         vis[u]=false;
         for(int i=head[u]; i!=-; 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]==-) return false;
     else return true;
 }
 int mincostflow(int s,int t,int &cost)
 {
     int flow=;
     cost=;
     while(spfa(s,t))
     {
         int Min=inf;
         for(int i=pre[t]; i!=-; i=pre[edge[i^].to]) //更新流量
         {
             if(Min>edge[i].cap-edge[i].flow)
                 Min=edge[i].cap-edge[i].flow;
         }
         for(int i=pre[t]; i!=-; i=pre[edge[i^].to])
         {
             edge[i].flow+=Min;
             edge[i^].flow-=Min;
             cost+=edge[i].cost*Min;
         }
         flow+=Min;
        // if(flow>=f)break;    //达到想要的f直接弹出来
     }
     return flow;
 }
 int main()
 {
     int n,m;
     while(~scanf("%d%d",&n,&m))
     {
         init(n+);
         for(int i=; i<m; i++)
         {
             int u,v,cap,cost;
             scanf("%d%d%d",&u,&v,&cost);
             addedge(u,v,,cost);
             addedge(v,u,,cost);
         }
         addedge(,,,);
         addedge(n,n+,,);
         int ans,maxflow;
         maxflow=mincostflow(,n+,ans);
         printf("%d\n",ans);
     }
 }