poj 2135 Farm Tour

本文介绍了一种解决特定路径寻找问题的算法——最短农场巡游问题。该问题要求找到从起点到终点并返回起点的最短路径,且每条路径只能经过一次。文章通过最小费用最大流算法实现,并详细展示了具体的实现代码。

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Farm Tour
Time Limit: 1000MS Memory Limit: 65536K
Total Submissions: 16367 Accepted: 6320
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
Source

USACO 2003 February Green


【分析】
最小费用最大流(费用流)板子…不过建图需要一定技巧
题意:给一个图,从1走到n,再从n走到1,且每条边只能走一边所走过的最小边权。

由于每条边只能走一次,所以给每条边设定流量为1,将点1与源点s连接,费用为0,流量为2,将n与汇点t连接,费用为0,流量为2。

跑一边费用流就好了。


【代码】

//poj 2135 Farm Tour 
#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define inf 1e9+7
#define M(a) memset(a,0,sizeof a)
#define fo(i,j,k) for(i=j;i<=k;i++)
using namespace std;
const int mxn=10005<<2;
queue <int> q;
int n,m,s,t,cnt,ans;
bool vis[mxn];
int head[1005],dis[1005],pre[1005];
struct node {int from,to,next,d,flow;} f[mxn];
inline void add(int u,int v,int d,int flow)
{
    f[++cnt].to=v,f[cnt].from=u,f[cnt].next=head[u],f[cnt].d=d,f[cnt].flow=flow,head[u]=cnt;
    f[++cnt].to=u,f[cnt].from=v,f[cnt].next=head[v],f[cnt].d=-d,f[cnt].flow=0,head[v]=cnt;
}
inline void spfa()
{
    int i,u,v,d,flow;
    memset(dis,0x3f,sizeof dis);
    memset(pre,-1,sizeof pre);
    M(vis);
    q.push(s);
    dis[s]=0;
    vis[s]=1;
    while(!q.empty())
    {
        u=q.front();
        q.pop();
        vis[u]=0;
        for(i=head[u];i;i=f[i].next)
        {
            v=f[i].to,d=f[i].d,flow=f[i].flow;
            if(dis[v]>dis[u]+d && flow>0)
            {
                dis[v]=dis[u]+d;
                pre[v]=i;   //记录前驱 
                if(!vis[v]) vis[v]=1,q.push(v);
            }
        }
    }
}
inline void maxflow()
{
    int i,u,v,d;
    spfa();
    while(pre[t]!=-1)
    {
        int tmp=inf;
        for(i=pre[t];i!=-1;i=pre[f[i].from])
          tmp=min(tmp,f[i].flow);
        ans+=dis[t]*tmp;
        for(i=pre[t];i!=-1;i=pre[f[i].from])
        {
            f[i].flow-=tmp;
            if(i&1) f[i+1].flow+=tmp;
            else f[i-1].flow+=tmp;
        }
        spfa();
    }
}
int main()
{
    int i,j,u,v,d;
    scanf("%d%d",&n,&m);
    s=0,t=n+1;
    while(m--)
    {
        scanf("%d%d%d",&u,&v,&d);
        add(u,v,d,1);add(v,u,d,1);
    }
    add(s,1,0,2),
    add(1,s,0,2),
    add(n,t,0,2),
    add(t,n,0,2);
    maxflow();
    printf("%d\n",ans);
    return 0;
}
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