Farm Tour

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

6

 

这个题因为还会返回所以需要建双向边，还会有重边问题，但是因为返回的路不可以和正向的路相同，所以建权值为w，容量为1的边，但是源点到1和n到汇点需要建容量为2，权值为0的边

#include <iostream>#include<stdio.h>#include<string.h>#include<stdlib.h>#include<math.h>#define oo 1<<28#include<queue>using namespace std;int pre[2100];int dis[2100];int vist[2100];int head[2100];struct node{    int u;    int v;//与u点相连的点    int f;//容量    int w;//权值    int next;//下一条边}edge[110000];int cnt=0;int s,t;void add(int u,int v,int f,int w){    edge[cnt].u=u;    edge[cnt].v=v;    edge[cnt].f=f;    edge[cnt].w=w;    edge[cnt].next=head[u];    head[u]=cnt++;    edge[cnt].u=v;    edge[cnt].v=u;    edge[cnt].f=0;    edge[cnt].w=-w;//反向的权值为正向的相反数    edge[cnt].next=head[v];    head[v]=cnt++;}void init(){    memset(pre,-1,sizeof(pre));    memset(vist,0,sizeof(vist));    for(int i=0;i<=t;i++)    {        dis[i]=oo;    }}int spfa(){    int i;    init();    queue<int>q;    dis[s]=0;    vist[s]=1;    q.push(s);    while(!q.empty())//将相邻点进行松弛，直到队列为空    {        int u=q.front();//取出队头        q.pop();        i=head[u];        vist[u]=0;        while(i!=-1)        {            int w=edge[i].w;            int v=edge[i].v;            if(edge[i].f>0&&dis[v]>dis[u]+w)//判断是否可以更新            {                dis[v]=dis[u]+w;//改进s到v点的值                pre[v]=i;//记录此点的前驱                if(!vist[v])//由于距离变小了，如果v点松弛成功且v点不在队列里，因为v点有可能还能改进别的点                {                    vist[v]=1;                    q.push(v);                }            }            i=edge[i].next;        }    }    if(pre[t]==-1)//如果不在最短路中代表寻找失败        return 0;    return 1;}int Cost(){    int ans=0;    while(spfa())//如果增广路寻找成功    {        int maxx=oo;        int p=pre[t];//初始化P指针        while(p!=-1)        {            maxx=min(edge[p].f,maxx);//求出此增广路上边的最小值            p=pre[edge[p].u];        }        p=pre[t];        while(p!=-1)        {            edge[p].f-=maxx;//正向减去            edge[p^1].f+=maxx;//反向增加            ans+=maxx*edge[p].w;//因为以单位计费，所以应当乘上流量            p=pre[edge[p].u];        }    }    return ans;}int main(){    int n,m,d;    int i,j;    while(~scanf("%d%d",&n,&m))    {        int u,v,w;        memset(head,-1,sizeof(head));        for(i=0;i<m;i++)        {            scanf("%d%d%d",&u,&v,&w);            add(u,v,1,w);            add(v,u,1,w);//之前没有看清楚，看了一下POJ别人给的注意发现是无向图        }        add(0,1,2,0);        add(n,n+1,2,0);//因为从源点到汇点和汇点到源点要走两遍，所以容量为2        s=0;t=n+1;        printf("%d\n",Cost());    }    return 0;}