hdu 1853 Cyclic Tour(费用流OR二分图最佳匹配，5级)

O - Cyclic Tour

Time Limit:1000MS     Memory Limit:65535KB     64bit IO Format:%I64d & %I64u

Submit  Status

System Crawler  (2013-05-30)

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

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
6 5
1 2 1
2 3 1
3 4 1
4 5 1
5 6 1

Sample Output

42 -1

Hint

In the first sample, there are two cycles, (1->2->3->1) and (6->5->4->6) whose length is 20 + 22 = 42.

#include<iostream>
#include<cstring>
#include<cstdio>
#include<queue>
#define FOR(i,a,b) for(int i=a;i<=b;++i)
#define clr(f,z) memset(f,z,sizeof(f))
using namespace std;
const int mm=234;
const int oo=1e9;
bool vis[mm];
int head[mm],edge,work[mm],dis[mm],src,dest,node;
class Edge
{
  public:int v,next,flow,cost;
}e[mm*mm*2];
void data()
{
  clr(head,-1);edge=0;
}
void add(int u,int v,int _flow,int _cost)
{
  e[edge].v=v;e[edge].flow=_flow;e[edge].cost=_cost;e[edge].next=head[u];head[u]=edge++;
  e[edge].v=u;e[edge].flow=0;e[edge].cost=-_cost;e[edge].next=head[v];head[v]=edge++;
}
bool spfa()
{
  FOR(i,0,node)dis[i]=oo,vis[i]=0;
  queue<int>Q;work[src]=work[dest]=-1;
  Q.push(src);int u,v;dis[src]=0;
  while(!Q.empty())
  {
    u=Q.front();Q.pop();vis[u]=0;
    for(int i=head[u];~i;i=e[i].next)
    {
      v=e[i].v;
      if(e[i].flow&&dis[v]>dis[u]+e[i].cost)
      {
        dis[v]=dis[u]+e[i].cost;work[v]=i^1;//path
        if(vis[v])continue;vis[v]=1;Q.push(v);
      }
    }
  }
  //FOR(i,0,node)printf("d=%d %d\n",i,dis[i]);
  return work[dest]!=-1;
}
int max_spfa(int& many)
{
  int ret=0,num;
  while(spfa())
  {
    num=oo;//puts("++");
    for(int i=work[dest];~i;i=work[e[i].v])
    {
      if(e[i^1].flow<num)num=e[i^1].flow;
    }
    ret+=num*dis[dest];many+=num;
    for(int i=work[dest];~i;i=work[e[i].v])
      e[i^1].flow-=num,e[i].flow+=num;
  }
  return ret;
}
int n,m;
int main()
{ int a,b,c;
  while(~scanf("%d%d",&n,&m))
  {
    data();src=0;dest=n+n+1;node=n+n+2;
    FOR(i,1,n)add(src,i,1,0),add(i+n,dest,1,0);
    FOR(i,1,m)
    {
      scanf("%d%d%d",&a,&b,&c);
      add(a,b+n,1,c);
    }
    int ret=0,ans;
    ans=max_spfa(ret);
    //cout<<ret<<endl;
    if(ret==n)
      printf("%d\n",ans);
    else printf("-1\n");
  }
  return 0;
}