Reverse graph of HDU2647 Reward (topological sorting)

Reverse graph of HDU2647 Reward (topological sorting)
RewardTime Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission (s): 4638 Accepted Submission (s): 1416


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 shoshould 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 shoshould 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 'demanders, print-1.
Sample Input

2 11 22 21 22 1

Sample Output

1777-1

Question: There are n people, and the minimum bonus for each person is 888. There are m requirements. m is a B, which means a requires B to have more bonuses. Ask if all the requirements can be met. If yes, the sum of the prizes received by n people is at least. If not, output-1.

#include
 
  #include
  
   #include
   
    #include
    
     using namespace std;const int N = 10005;int n,mony[N],in[N];vector
     
      mapt[N];void init(){    for(int i=1;i<=n;i++)        mony[i]=888,in[i]=0,mapt[i].clear();}int tope(){    queue
      
       q; int k=0,sum=0; for(int i=1;i<=n;i++) if(in[i]==0) { k++; q.push(i); in[i]=-1; sum+=mony[i]; } while(!q.empty()) { int s=q.front(); q.pop(); for(int i=0;i
       
        0) { in[j]--; if(mony[j]
        
         0) { init(); while(m--) { scanf("%d%d",&a,&b); in[a]++; mapt[b].push_back(a); } printf("%d\n",tope()); }}