拓扑&&reward

http://acm.hdu.edu.cn/showproblem.php?pid=2647

Problem Description

Dandelion's uncle is a boss of a factory. As the spring festival is coming , he wants to distribute rewards to his workers. Now he has a trouble about how to distribute the rewards.
The workers will compare their rewards ,and some one may have demands of the distributing of rewards ,just like a's reward should more than b's.Dandelion's unclue wants to fulfill all the demands, of course ,he wants to use the least money.Every work's reward will be at least 888 , because it's a lucky number.

 

Input

One line with two integers n and m ,stands for the number of works and the number of demands .(n<=10000,m<=20000)
then m lines ,each line contains two integers a and b ,stands for a's reward should be more than b's.

 

Output

For every case ,print the least money dandelion 's uncle needs to distribute .If it's impossible to fulfill all the works' demands ,print -1.

 

Sample Input

   
   
   
   

    
    
    
    2 1 1 2 2 2 1 2 2 1
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    1777 -1
   
   
   
   

这个题是简单的拓扑排序。。。以前做过，这次就是把它重新回忆一下。。

代码：

#include<iostream>
#include<vector>
#include<queue>
#include<algorithm>
using namespace std;
#define N 10005
int in[N];
int ran[N];
vector<int> v[N];
int n,m;
bool tp()
{   queue<int> q;
    for(int i=1;i<=n;i++)
      if(in[i]==0)
       q.push(i);
       int count=0;
         while(!q.empty())
         {  count++;
           int t=q.front();
             q.pop();
            for(int i=0;i<v[t].size();i++ )
              { int k=v[t][i];
              ran[k]=max(ran[k],ran[t]+1);//记录k节点有多少前继节点，为得到至少得到的工资数。
              if(in[k]==1) q.push(k);
              else in[k]--;
              }
         }
         if(count==n) return true;//判断是否形成环。
          else return false;
}
int main()
{    while(cin>>n>>m)
    {    for(int i=0;i<=n;i++)
          { v[i].clear();
            in[i]=0;
            ran[i]=0;
          }
           for(int i=0;i!=m;i++)
           {  int a,b;
                cin>>a>>b;
                v[b].push_back(a);//把后继点存在本节点的邻接表内。
                  in[a]++;//记录每一点的入度。
           }
           if(tp())
           { int ans=0;
           for(int i=1;i<=n;i++)
              ans+=ran[i];
               cout<<n*888+ans<<endl;
           }
            else cout<<"-1"<<endl;     
  }return 0;
    
}