【杭电oj2647】Reward

                                                   Reward

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7528    Accepted Submission(s): 2377

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 11 22 21 22 1

 

Sample Output

1777-1

 

Author

dandelion

 

Source

曾是惊鸿照影来

 

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数据太大用邻接表+队列，每一层薪水加一块钱（这老板真抠），上一层加一即可。

#include<stdio.h>  #include<string.h>  #include<queue>  #include<algorithm>  using namespace std;  struct zz  {      int to;      int next;  }p[20200];  int in[10100];  int head[10100];  int num[10100];  void topo(int n)  {      queue<int>q;      int i,j,k;      int ans=0;      int sum=0;      for(i=1;i<=n;i++)      {          if(in[i]==0)          q.push(i);      }      while(!q.empty())      {          k=q.front();          q.pop();          ans++;          for(j=head[k];j!=-1;j=p[j].next)          {              in[p[j].to]--;              if(in[p[j].to]==0)              {                  q.push(p[j].to);                  num[p[j].to]=num[k]+1;  //重点看懂这一步            }          }      }      for(i=1;i<=n;i++)          sum+=num[i];          if(ans<n)              printf("-1\n");          else              printf("%d\n",sum);  }  int main(){      int n,m,i,j,a,b;      while(scanf("%d%d",&n,&m)!=EOF)      {          memset(in,0,sizeof(in));          memset(head,-1,sizeof(head));          for(i=1;i<=n;i++)              num[i]=888;          for(i=0;i<m;i++)          {              scanf("%d%d",&a,&b);              p[i].to=a;              p[i].next=head[b];              head[b]=i;              in[a]++;          }          topo(n);      }      return 0;  }