Cyclic Tour

Problem Description

There are N cities in our country, and M one-way roads connecting them. Now Little Tom wants to make several cyclic tours, which satisfy that, each cycle contain at least two cities, and each city belongs to one cycle exactly. Tom wants the total length of all the tours minimum, but he is too lazy to calculate. Can you help him?

 

Input

There are several test cases in the input. You should process to the end of file (EOF).
The first line of each test case contains two integers N (N ≤ 100) and M, indicating the number of cities and the number of roads. The M lines followed, each of them contains three numbers A, B, and C, indicating that there is a road from city A to city B, whose length is C. (1 ≤ A,B ≤ N, A ≠ B, 1 ≤ C ≤ 1000).

 

Output

Output one number for each test case, indicating the minimum length of all the tours. If there are no such tours, output -1.

 

Sample Input

 

Sample Output

 

Author

RoBa@TJU

 

Source

HDU 2007 Programming Contest - Final

 

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#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
using namespace std;
#define MAXN 210
#define INF 0x3f3f3f3f
//AC
int n,m;
struct Edge 
{
        int np,val,c;
        Edge *next,*neg;
}E[MAXN*MAXN],*V[MAXN*2];
int tope,sour=0,sink=1;
void add_edge(int x,int y,int z,int c)
{
        //cout<<"Add"<<x<<" "<<y<<" "<<z<<" "<<c<<endl;
        E[++tope].np=y;
        E[tope].val=z;
        E[tope].c=c;
        E[tope].next=V[x];
        V[x]=&E[tope];
        E[++tope].np=x;
        E[tope].val=0;
        E[tope].c=-c;
        E[tope].next=V[y];
        V[y]=&E[tope];

        E[tope].neg=&E[tope-1];
        E[tope-1].neg=&E[tope];
}
int q[MAXN*10],vis[MAXN],dis[MAXN],dfn=0;
int prev[MAXN];
Edge *path[MAXN];
int spfa()
{
        int ope=-1,clo=0,now;
        Edge *ne;
        memset(dis,INF,sizeof(dis));
        dfn++;
        q[0]=sour;
        vis[sour]=dfn;
        dis[sour]=0;
        while (ope<clo)
        {
                now=q[++ope];
                vis[now]=0;
                for (ne=V[now];ne;ne=ne->next)
                {
                        if (ne->val&&dis[ne->np]>dis[now]+ne->c)
                        {
                                dis[ne->np]=dis[now]+ne->c;
                                prev[ne->np]=now;
                                path[ne->np]=ne;
                                if (vis[ne->np]!=dfn)
                                {
                                        vis[ne->np]=dfn;
                                        q[++clo]=ne->np;
                                }
                        }
                }
        }
        return dis[sink];
}
pair<int,int> max_cost_flow()
{
        int ds,fl,now,x;
        pair<int,int> ret;
        ret.first=ret.second=0;
        while (ds=spfa(),ds!=INF)
        {
                x=sink;
                fl=INF;
                while (x!=sour)
                {
                        fl=min(fl,path[x]->val);
                        x=prev[x];
                }
                x=sink;
                while (x!=sour)
                {
                        path[x]->val-=fl;
                        path[x]->neg->val+=fl;
                        x=prev[x];
                }
                ret.first+=fl;
                ret.second+=ds*fl;
        }
        return ret;
}
int main()
{
        freopen("input.txt","r",stdin);
        int i,j,k,x,y,z;
        while (~scanf("%d%d",&n,&m))
        {
                tope=-1;
                memset(V,0,sizeof(V));
                for (i=0;i<n;i++)
                {
                        add_edge(sour,2+i,1,0);
                        add_edge(2+i+n,sink,1,0);
                }
                for(i=0;i<m;i++)
                {
                        scanf("%d%d%d",&x,&y,&z);x--;y--;
                        add_edge(2+x,2+y+n,1,z);
                }
                pair<int,int> p1;
                p1=max_cost_flow();
                if (p1.first!=n)
                {
                        printf("-1\n");
                }else
                {
                        printf("%d\n",p1.second);
                }
        }
}