hdu 3488 Tour HDU （费用流）

In the kingdom of Henryy, there are N (2 <= N <= 200) cities, with M (M <= 30000) one-way roads connecting them. You are lucky enough to have a chance to have a tour in the kingdom. The route should be designed as: The route should contain one or more loops. (A loop is a route like: A->B->……->P->A.) 
Every city should be just in one route. 
A loop should have at least two cities. In one route, each city should be visited just once. (The only exception is that the first and the last city should be the same and this city is visited twice.) 
The total distance the N roads you have chosen should be minimized. 

Input

An integer T in the first line indicates the number of the test cases. 
In each test case, the first line contains two integers N and M, indicating the number of the cities and the one-way roads. Then M lines followed, each line has three integers U, V and W (0 < W <= 10000), indicating that there is a road from U to V, with the distance of W. 
It is guaranteed that at least one valid arrangement of the tour is existed. 
A blank line is followed after each test case.

Output

For each test case, output a line with exactly one integer, which is the minimum total distance.

Sample Input

1
6 9
1 2 5
2 3 5
3 1 10
3 4 12
4 1 8
4 6 11
5 4 7
5 6 9
6 5 4

Sample Output

42

2、最开始看到需要走一个圈的时候可能稍微有点绕，不知道从何下手比较好，其实并不难，首先我们对应想到如果从一个点出发，再回到这个点，如果对应有流需要流过的话，其实对应我们可以将点i拆成两个点i，i+n，一个用来表示起点，一个用来表示终点。

3、那么问题转化到费用流上来：

①建立源点S，将源点连入各个节点（1-n），费用为0，容量为1.表示一个点只能走一次。

②建立汇点T，将各个拆点连入汇点（n+1-2n），费用为0，容量为1。

③对应m条有向边，将u连入v+n，费用为其权值，容量为INF。

那么对应一个简单的三个点组成的环的建图情况：

很明显，跑完费用流的maxflow=3；

那么如果少了3->1这条有向边，很明显maxlfow=2；

4、那么我们得到结论：建好图之后跑费用流，因为题目中保证了一定有解，那么直接输出最小花费即可。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <cmath>
#include <vector>
#include <queue>
#include <algorithm>
using namespace std;
const int N = 1e3+10;
const int inf = 0x3f3f3f3f;
typedef pair<int,int>P;
char str[N];
struct node
{
    int x, y;
} v1[N], v2[N];

struct edge
{
    int to, cap, cost, rev;
};
vector<edge>p[N];
void add(int u,int v,int cap,int w)
{
    edge e;
    e.to=v, e.cap=cap, e.cost=w, e.rev=p[v].size();
    p[u].push_back(e);
    e.to=u, e.cap=0, e.cost=-w, e.rev=p[u].size()-1;
    p[v].push_back(e);
    return ;
}

int d[N], h[N], prev1[N], pree[N];

int min_cost_flow(int s,int t,int f)
{
    int res=0;
    memset(h,0,sizeof(h));
    while(f>0)
    {
        fill(d,d+t+1,inf);
        priority_queue< P, vector<P>, greater<P> >q;
        q.push(P(0,s));
        d[s]=0;
        while(!q.empty())
        {
            P x=q.top();
            q.pop();
            int w=x.first, u=x.second;
            if(d[u]<w) continue;
            for(int i=0; i<p[u].size(); i++)
            {
                int v=p[u][i].to;
                w=p[u][i].cost;
                if(p[u][i].cap>0&&d[v]>d[u]+w+h[u]-h[v])
                {
                    d[v]=d[u]+w+h[u]-h[v];
                    prev1[v]=u, pree[v]=i;
                    q.push(P(d[v],v));
                }
            }
        }
        if(d[t]==inf) return -1;
        for(int i=0; i<=t; i++) h[i]+=d[i];
        int flow=f;
        for(int i=t; i!=s; i=prev1[i])
        {
            flow=min(flow,p[prev1[i]][pree[i]].cap);
        }
        f-=flow, res+=(h[t]*flow);
        for(int i=t; i!=s; i=prev1[i])
        {
            edge &e=p[prev1[i]][pree[i]];
            e.cap-=flow;
            p[e.to][e.rev].cap+=flow;
        }
    }
    return res;
}

int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        int n, m;
        scanf("%d %d", &n, &m);
        int s=0, e=2*n+1;
        for(int i=1; i<=n; i++)
            add(s,i,1,0),add(n+i,e,1,0);
        for(int i=0; i<m; i++)
        {
            int x, y, z;
            scanf("%d %d %d", &x, &y, &z);
            add(x,n+y,inf,z);
        }
        int x=min_cost_flow(s, e, n);
        printf("%d\n",x);
        for(int i=0; i<=e; i++) p[i].clear();
    }
    return 0;
}