Farm Tour(费用流模板题)

最新推荐文章于 2020-09-13 23:41:39 发布

转载最新推荐文章于 2020-09-13 23:41:39 发布 · 129 阅读

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原文链接：http://www.cnblogs.com/nzher/p/6581005.html

本文介绍了一个关于寻找农场中从家到谷仓再返回的最短路径问题，并提供了一种解决该问题的具体算法实现。该问题涉及图论中的路径搜索与优化，通过建立图模型并利用特定算法求解最短路径。

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题目描述

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.

输入

* 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.

输出

A single line containing the length of the shortest tour.

样例输入

样例输出

#include<cstdio>  
#include<cstring>  
struct edge{  
    int to,cap,rev,nx,we;  
}G[40050];  
int n,m,p;  
int h[2050],q[40050],d[2050];  
bool mark[2050];  
int Min(int a,int b){return a<b?a:b;}  
void ae(int s,int e,int c,int w){  
    G[++p]=(edge){e,c,p+1,h[s],w};h[s]=p;  
    G[++p]=(edge){s,0,p-1,h[e],-w};h[e]=p;  
}  
bool spfa(){  
    memset(d,127/3,sizeof(d));  
    memset(mark,0,sizeof(mark));  
    d[n+1]=0;mark[n+1]=1;  
    int head=0,tail=0;  
    int inf=d[0];  
    q[tail++]=n+1;  
    while(head!=tail){  
        int fr=q[head++];  
        for(int i=h[fr];i;i=G[i].nx){  
            if(G[G[i].rev].cap>0&&d[G[i].to]>d[fr]+G[G[i].rev].we){  
                d[G[i].to]=d[fr]+G[G[i].rev].we;  
                if(!mark[G[i].to])mark[G[i].to]=1,q[tail++]=G[i].to;  
            }  
        }  
        mark[fr]=0;  
    }  
    return !(d[0]==inf);  
}  
int dfs(int s,int t,int f){  
    mark[s]=1;  
    if(s==t) return f;  
    int sum=0;  
    for(int i=h[s];i;i=G[i].nx){  
        if(G[i].cap>0&&!mark[G[i].to]&&d[s]-G[i].we==d[G[i].to]){  
            int d=dfs(G[i].to,t,Min(G[i].cap,f));  
            if(d)sum+=d,f-=d,G[i].cap-=d,G[G[i].rev].cap+=d;  
            if(f==0)return sum;  
        }  
    }  
    return sum;  
}  
int m__f(){  
    int sum=0;  
    while(spfa()){  
        mark[n+1]=1;  
        while(mark[n+1]){  
            memset(mark,0,sizeof(mark));  
            sum+=dfs(0,n+1,99999999)*d[0];  
        }  
    }  
    return sum;  
}  
int main(){  
    scanf("%d%d",&n,&m);  
    for(int i=1;i<=m;i++){  
        int x,y,z;  
        scanf("%d%d%d",&x,&y,&z);  
        ae(x,y,1,z);ae(y,x,1,z);  
    }  
    ae(0,1,2,0);ae(n,n+1,2,0);  
    printf("%d\n",m__f());  
    return 0;  
}