HDU3488 Cyclic Tour(有向环最小覆盖)

思路：和hdu1853几乎一样的代码和思路...

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
#include<vector>
#define INF 1e9
using namespace std;
const int maxn=400+5;

struct Edge
{
    int from,to,cap,flow,cost;
    Edge(){}
    Edge(int f,int t,int c,int fl,int co):from(f),to(t),cap(c),flow(fl),cost(co){}
};

struct MCMF
{
    int n,m,s,t;
    vector<Edge> edges;
    vector<int> G[maxn];
    bool inq[maxn];
    int d[maxn];
    int p[maxn];
    int a[maxn];

    void init(int n,int s,int t)
    {
        this->n=n, this->s=s, this->t=t;
        edges.clear();
        for(int i=0;i<n;++i) G[i].clear();
    }

    void AddEdge(int from,int to,int cap,int cost)
    {
        edges.push_back(Edge(from,to,cap,0,cost));
        edges.push_back(Edge(to,from,0,0,-cost));
        m=edges.size();
        G[from].push_back(m-2);
        G[to].push_back(m-1);
    }

    bool BellmanFord(int &flow,int &cost)
    {
        queue<int> Q;
        for(int i=0;i<n;++i) d[i]=INF;
        memset(inq,0,sizeof(inq));
        Q.push(s),inq[s]=true,d[s]=0,a[s]=INF,p[s]=0;

        while(!Q.empty())
        {
            int u=Q.front(); Q.pop();
            inq[u]=false;
            for(int i=0;i<G[u].size();++i)
            {
                Edge &e=edges[G[u][i]];
                if(e.cap>e.flow && d[e.to]>d[u]+e.cost)
                {
                    d[e.to]=d[u]+e.cost;
                    a[e.to]=min(a[u],e.cap-e.flow);
                    p[e.to]=G[u][i];
                    if(!inq[e.to]){inq[e.to]=true; Q.push(e.to);}
                }
            }
        }
        if(d[t]==INF) return false;
        flow += a[t];
        cost += a[t]*d[t];
        int u=t;
        while(u!=s)
        {
            edges[p[u]].flow +=a[t];
            edges[p[u]^1].flow -=a[t];
            u=edges[p[u]].from;
        }
        return true;
    }

    int solve(int num)
    {
        int flow=0,cost=0;
        while(BellmanFord(flow,cost));
        return flow==num?cost:-1;
    }
}MM;

int main()
{
    int T;scanf("%d",&T);
    while(T--)
    {
        int n,m;
        scanf("%d%d",&n,&m);
        int src=0,dst=2*n+1;
        MM.init(2*n+2,src,dst);
        for(int i=1;i<=n;++i)
        {
            MM.AddEdge(src,i,1,0);
            MM.AddEdge(i+n,dst,1,0);
        }
        while(m--)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            MM.AddEdge(u,v+n,1,w);
        }
        printf("%d\n",MM.solve(n));
    }
    return 0;
}

Description

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