【poj2135】Farm Tour

裸的費用流
源點向家引容量爲2的邊
穀倉向匯點引容量爲2的邊
其餘的按照原圖建圖，容量爲1，費用爲長度
等於一共跑了兩次
直接跑費用流.

#include<iostream>
#include<cstdio>
#include<cstring>
#include<string>
#include<algorithm>
#include<queue>

using namespace std;
const int N=10000,inf=0x3f3f3f3f;
int d[N],p[N],a[N],head[N],inq[N];
int te,s,t,n,m;
queue<int>q;
struct edge{
    int u,v,cap,cost,flow,next;
}e[100010];
void add(int u,int v,int cap,int cost)
{
    e[++te].u=u;
    e[te].v=v;
    e[te].cap=cap;
    e[te].cost=cost;
    e[te].next=head[u];
    head[u]=te;
}
void insert(int u,int v,int cap,int cost)
{
    add(u,v,cap,cost),add(v,u,0,-cost);
}
int spfa(int &flow,int &cost)
{
    memset(a,0,sizeof(a));
    memset(d,0x3f,sizeof(d));
    memset(inq,0,sizeof(inq));
    while(!q.empty())q.pop();
    q.push(s);d[s]=0;inq[s]=1,a[s]=inf;
    while(!q.empty())
    {
        int u=q.front();
        for (int i=head[u];i;i=e[i].next)
        {
            int v=e[i].v;
            if (e[i].cap>e[i].flow&&d[v]>d[u]+e[i].cost)
            {
                d[v]=d[u]+e[i].cost;
                a[v]=min(e[i].cap-e[i].flow,a[u]);
                p[v]=i;
                if (!inq[v])
                {
                    inq[v]=1;
                    q.push(v);
                }
            }
        }
        inq[u]=0;
        q.pop();
    }
    if (d[t]==inf)return false;
    cost+=d[t]*a[t];
    flow+=a[t];
    for (int u=t;u!=s;u=e[p[u]].u)
    {
        e[p[u]].flow+=a[t];
        e[p[u]^1].flow-=a[t];
    }
    return true;
}
int mcmf(int &cost)
{
    int flow=0;
    cost=0;
    while(spfa(flow,cost));
    return flow;
}
int main()
{
    cin>>n>>m;
    int u,v,cost;
    te=1;s=(n<<1)+1,t=(n<<1)+2;
    insert(s,1,2,0);insert(n,t,2,0);
    for (int i=1;i<=m;++i)
    {
        scanf("%d%d%d",&u,&v,&cost);
        insert(u,v,1,cost);
        insert(v,u,1,cost);
    }
    mcmf(cost);
    cout<<cost;
}