poj 2135(最小费用最大流)

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 5
1 2 1
2 3 1
3 4 1
1 3 2
2 4 2

Sample Output

6

Source

USACO 2003 February Green

题意:农夫想从 1 走到 n,然后又从 n 走到 1，要求两次所走的路径不一样并且和最小,问最小路径和?

题解:从题目中我们可以知道这个题的最大流为2，所以添加容量为2的源点和汇点进行限流,然后再将边的权值变成费用，跑一遍最小费用最大流即可。

#include <cstdio>
#include <cstring>
#include <queue>
#include <algorithm>
using namespace std;
const int INF = 999999999;
const int N = 2005;
const int M = 80005;
struct Edge{
    int u,v,cap,cost,next;
}edge[M];
int head[N],tot,low[N],pre[N];
int total ;
bool vis[N];
void addEdge(int u,int v,int cap,int cost,int &k){
    edge[k].u=u,edge[k].v=v,edge[k].cap = cap,edge[k].cost = cost,edge[k].next = head[u],head[u] = k++;
    edge[k].u=v,edge[k].v=u,edge[k].cap = 0,edge[k].cost = -cost,edge[k].next = head[v],head[v] = k++;
}
void init(){
    memset(head,-1,sizeof(head));
    tot = 0;
}
bool spfa(int s,int t,int n){
    memset(vis,false,sizeof(vis));
    for(int i=0;i<=n;i++){
        low[i] = (i==s)?0:INF;
        pre[i] = -1;
    }
    queue<int> q;
    q.push(s);
    while(!q.empty()){
        int u = q.front();
        q.pop();
        vis[u] = false;
        for(int k=head[u];k!=-1;k=edge[k].next){
            int v = edge[k].v;
            if(edge[k].cap>0&&low[v]>low[u]+edge[k].cost){
                low[v] = low[u] + edge[k].cost;
                pre[v] = k; ///v为终点对应的边
                if(!vis[v]){
                    vis[v] = true;
                    q.push(v);
                }
            }
        }
    }
    if(pre[t]==-1) return false;
    return true;
}
int MCMF(int s,int t,int n){
    int mincost = 0,minflow,flow=0;
     while(spfa(s,t,n))
    {
        minflow=INF+1;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
            minflow=min(minflow,edge[i].cap);
        flow+=minflow;
        for(int i=pre[t];i!=-1;i=pre[edge[i].u])
        {
            edge[i].cap-=minflow;
            edge[i^1].cap+=minflow;
        }
        mincost+=low[t]*minflow;
    }
    total=flow;
    return mincost;
}
int n,m;

int main(){
    while(scanf("%d%d",&n,&m)!=EOF){
        init();
        int src = 0,des = n+1;
        for(int i=1;i<=m;i++){
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            if(u==v) continue;
            addEdge(u,v,1,w,tot);
            addEdge(v,u,1,w,tot);
        }
        addEdge(src,1,2,0,tot);
        addEdge(n,des,2,0,tot);
        int a = MCMF(src,des,n+2);
        printf("%d\n",a);
    }
}