费用流！

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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 51 2 12 3 13 4 11 3 22 4 2

Sample Output

—

—基本思路：
把弧<i,j>的单位费用w[i,j]看作弧<i,j>的路径长度，每次找从源点s到汇点t长度最短（费用最小）的可增广路径进行增广。
1. 最小费用可增广路
2. 路径s到t的长度即单位流量的费用。

其实就是把ek算法中的广搜改成spfa……（因为存在负权边所以得用bellman-ford或spfa）

<pre name="code" class="cpp">#include"stdio.h"#include"string.h"#include"queue"using namespace std;#define M 1100const int inf=0x7fffffff;struct node{    int u,v,c,f,next;  //C为花费，F为flow流量}e[M*40];int pre[M],dis[M],head[M],t;int vis[M];void add1(int u,int v,int c,int f){    e[t].u=u;    e[t].v=v;    e[t].c=c;    e[t].f=f;    e[t].next=head[u];    head[u]=t++;}void add(int u,int v,int c,int f){    add1(u,v,c,f);          add1(v,u,-c,0);  //反向边流量初始为零，如果走反向边费用正好和原边抵消}int spfa(int s,int t){    int i,u,v;    queue<int>q;    q.push(s);    memset(vis,0,sizeof(vis));    memset(pre,-1,sizeof(pre));    for(i=s;i<=t;i++)        dis[i]=inf;    dis[s]=0;    while(!q.empty())    {        u=q.front();        q.pop();        for(i=head[u];i!=-1;i=e[i].next)        {            v=e[i].v;            if(e[i].f&&dis[v]>dis[u]+e[i].c)  //找到一条最小费用流            {                dis[v]=dis[u]+e[i].c;                  pre[v]=i;       //记录路径                if(!vis[v])                {                    vis[v]=1;                    q.push(v);                }            }        }        vis[u]=0;    }    if(dis[t]!=inf)        return 1;    return 0;}void solve(int s,int t){    int ans=0,i,j;int flow=0,cost=0;     //总流量、总费用    while(spfa(s,t))    {int minf=inf;for(i=pre[t];i!=-1;i=pre[e[i].u]){if(e[i].f<minf)minf=e[i].f;}flow+=minf;   //该条路径的流量        for(i=pre[t];i!=-1;i=pre[e[i].u])        {            j=i^1;            e[i].f-=minf;            e[j].f+=minf;        }cost+=dis[t]*minf;   //单位运费和乘以流量得费用       }    printf("%d\n",cost);}int main(){    int i,u,v,c,n,m;    while(scanf("%d%d",&n,&m)!=-1)    {        t=0;        memset(head,-1,sizeof(head));        for(i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&c);  //边长度看做单位运费            add(u,v,c,1);            add(v,u,c,1);       //无向边，费用为长度，流量为1（只能通过一次）        }//往返的两条线路可以看成是从源点到汇点的两条线路,//所以只要从附加源点向源点连一条容量为2的边和从汇点向附加汇点连一条容量为2的边就可以限线路为两条.        add(0,1,0,2);        add(n,n+1,0,2);        solve(0,n+1);    }    return 0;}

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