hdu1853 最小费用流

Cyclic Tour

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/65535 K (Java/Others)
Total Submission(s): 914    Accepted Submission(s): 466

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

   
   
   
   

    
    
    
    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. 
    
    
    
    
     
   
   
   
   

 

Author

RoBa@TJU

 

Source

HDU 2007 Programming Contest - Final

 

Recommend

lcy

 

第一次做费用流~~WA了几次~~~数组要大~~

思路：

环:则入度==出度

拆点建图，费用是长度。

感觉好新鲜。。

#include <iostream>
#include <cstring>
#include <cstdio>
#include <queue>
#define inf 1000000000
using namespace std;
struct node
{
    int u,v,flow,cost;
};
node e[50100];
int first[505],next[50100],preedge[505],p[505],d[505];
int ans_flow=0,ans_cost=0;
//int cc;
void add_edge(int u,int v,int f,int c,int &cc)
{
    e[cc].u=u;
    e[cc].v=v;
    e[cc].flow=f;
    e[cc].cost=c;
    next[cc]=first[u];
    first[u]=cc;
    cc++;

    e[cc].u=v;
    e[cc].v=u;
    e[cc].flow=0;
    e[cc].cost=-c;
    next[cc]=first[v];
    first[v]=cc;
    cc++;
}
bool spfa(int s,int t)
{
    int inq[205];
    memset(inq,0,sizeof(inq));
    memset(p,-1,sizeof(p));
    memset(preedge,-1,sizeof(preedge));
    int i;
    for(i=0;i<=t;i++)
        d[i]=inf;
    d[0]=0;
    queue<int> q;
    q.push(s);
    while(!q.empty())
    {
        int u=q.front();
        q.pop();
        inq[u]=0;
        for(int i=first[u];i!=-1;i=next[i])
        {
            if(e[i].flow)
            {
                int v=e[i].v;
                if(d[v]>d[u]+e[i].cost)
                {
                    p[v]=u;
                    preedge[v]=i;
                    d[v]=d[u]+e[i].cost;
                    if(!inq[v])
                    {
                        inq[v]=1;
                        q.push(v);
                    }
                }
            }
        }
    }
    if(d[t]>=inf)
        return false;
    else
        return true;
}
void min_cost_flow(int s,int t)
{
    int u;
    ans_flow=0,ans_cost=0;
    while(spfa(s,t))
    {
        u=t;
        int mm=inf;
        while(p[u]!=-1)
        {
            mm=min(mm,e[preedge[u]].flow);
            u=p[u];
        }
        u=t;
        while(p[u]!=-1)
        {
            e[preedge[u]].flow-=mm;
            e[preedge[u]^1].flow+=mm;
            u=p[u];
        }
        ans_cost+=d[t]*mm;
        ans_flow+=mm;
    }
}
int main()
{
    int n,m;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int i;
        memset(first,-1,sizeof(first));
        memset(next,-1,sizeof(next));
        memset(e,0,sizeof(e));
        int s=0;
        int t=2*n+1;
        int cc=0;
        for(i=0;i<m;i++)
        {
            int u,v,w;
            scanf("%d%d%d",&u,&v,&w);
            add_edge(u,n+v,1,w,cc);
        }
        for(i=1;i<=n;i++)
        {
            add_edge(0,i,1,0,cc);
            add_edge(n+i,t,1,0,cc);
        }
        min_cost_flow(s,t);
        //printf("%d\n",ans_flow);
        if(ans_flow==n)
            printf("%d\n",ans_cost);
        else
            printf("-1\n");
    }
    return 0;
}