POJ 2135(Farm Tour-费用流)[Template:费用流 V2]-优快云博客

本文介绍了一种解决农场导览最短路径问题的方法，利用最小费用流算法找到从农场主的房子到谷仓再返回的最短路线，确保每条路径只经过一次。

Language:

Farm Tour

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 11916		Accepted: 4454

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

Sample Output

Source

USACO 2003 February Green

直接求从S到T,容量为2的最小费用流。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<functional>
#include<iostream>
#include<cmath>
#include<cctype>
#include<ctime>
using namespace std;
#define For(i,n) for(int i=1;i<=n;i++)
#define Fork(i,k,n) for(int i=k;i<=n;i++)
#define Rep(i,n) for(int i=0;i<n;i++)
#define ForD(i,n) for(int i=n;i;i--)
#define RepD(i,n) for(int i=n;i>=0;i--)
#define Forp(x) for(int p=pre[x];p;p=next[p])
#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  
#define Lson (x<<1)
#define Rson ((x<<1)+1)
#define MEM(a) memset(a,0,sizeof(a));
#define MEMI(a) memset(a,127,sizeof(a));
#define MEMi(a) memset(a,128,sizeof(a));
#define INF (2139062143)
#define F (100000007)
#define MAXN (1000+10)
#define MAXM (10000*4+10)
#define MAXAi (35000)
#define eps (1e-3)
long long mul(long long a,long long b){return (a*b)%F;}
long long add(long long a,long long b){return (a+b)%F;}
long long sub(long long a,long long b){return (a-b+(a-b)/F*F+F)%F;}
typedef long long ll;

class Cost_Flow  
{  
public:  
    int n,s,t;  
    int q[10000];  
    int edge[MAXM],next[MAXM],pre[MAXN],weight[MAXM],size;  
    double cost[MAXM];  
    void addedge(int u,int v,int w,double c)    
    {    
        edge[++size]=v;    
        weight[size]=w;    
        cost[size]=c;    
        next[size]=pre[u];    
        pre[u]=size;    
    }    
    void addedge2(int u,int v,int w,double c){addedge(u,v,w,c),addedge(v,u,0,-c);}   
    bool b[MAXN];  
    double d[MAXN];  
    int pr[MAXN],ed[MAXN];  
    bool SPFA(int s,int t)    
    {    
        For(i,n) d[i]=INF;  
        MEM(b)  
        d[q[1]=s]=0;b[s]=1;    
        int head=1,tail=1;    
        while (head<=tail)    
        {    
            int now=q[head++];    
            Forp(now)    
            {    
                int &v=edge[p];    
                if (weight[p]&&d[now]+cost[p]<d[v])    
                {    
                    d[v]=d[now]+cost[p];    
                    if (!b[v]) b[v]=1,q[++tail]=v;    
                    pr[v]=now,ed[v]=p;    
                }    
            }    
            b[now]=0;    
        }    
        return fabs(d[t]-INF)>eps;    
    }   
    double totcost;    
          
    double CostFlow(int s,int t)    
    {    
        while (SPFA(s,t))    
        {    
            int flow=INF;    
            for(int x=t;x^s;x=pr[x]) flow=min(flow,weight[ed[x]]);      
            totcost+=(double)flow*d[t];    
            for(int x=t;x^s;x=pr[x]) weight[ed[x]]-=flow,weight[ed[x]^1]+=flow;         
        }    
        return totcost;    
    }    
    void mem(int n,int t)  
    {  
        (*this).n=n;  
        size=1;  
        totcost=0;  
        MEM(pre) MEM(next)   
    }  
}S;  
int n,m;
int main()
{
//	freopen("poj2135.in","r",stdin);
//	freopen(".out","w",stdout);
	cin>>n>>m; 
	S.mem(n+1,n);
	int s=n+1;
	S.addedge2(s,1,2,0);
	For(i,m)
	{
		int u,v,c;
		scanf("%d%d%d",&u,&v,&c);
		S.addedge2(u,v,1,c);
		S.addedge2(v,u,1,c);
	}
	
	cout<<(int)round(S.CostFlow(n+1,n))<<endl;
	
	return 0;
}