杭电2647 Reward(拓扑排序)

解决奖励分配问题的算法

最新推荐文章于 2018-09-16 19:22:17 发布

原创最新推荐文章于 2018-09-16 19:22:17 发布 · 681 阅读

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本文探讨了在有限预算下如何最优地分配奖励给工人，同时满足特定的奖励比较需求。通过建立图模型并应用拓扑排序算法，解决了奖励分配问题，并通过实例展示了算法的应用。

Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4099 Accepted Submission(s): 1244

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

Sample Output

1777
-1/*
一个错误搞了我2天。。。。
相当于将其分层，求层数
注意：判断是否有环是通过节点数与出队的总数判断的，此题输入要注意前后顺序，此题输入将a、b是将a的入度+1，不能G[a].push_back(b);
Time:2014-8-7 12:08
*/
#include<queue>
#include<stdio.h>
#include<string.h>
using namespace std;
const int MAX=10000+10;
int In[MAX],salary[MAX];
int n,m;
vector<vector<int > >G(MAX);
void Topsort()
{
     queue<int >q;
     int num=n;
     for(int i=1;i<=n;i++)
     {
      if(In[i]==0){q.push(i);In[i]=-1;}//入队后的点可以不标记为-1 In因为只检索相领接的后边的 
     }
     while(!q.empty())
     {                
        num--;
        int cur=q.front();
        q.pop();
        for(int i=0;i<G[cur].size();i++)
        {
          int temp=G[cur][i];
              In[temp]--;    
          if(In[temp]==0)
          {q.push(temp);salary[temp]=salary[cur]+1;In[temp]=-1;}
        }          
          
          //printf("%d\n",ans);  
     }
     if(num>0)
     printf("-1\n");
     else
     { 
         int summon=n*888;
      for(int i=1;i<=n;i++)
      summon+=salary[i];
      printf("%d\n",summon);
      }
}
int main()
{
    
    while(scanf("%d%d",&n,&m)!=EOF)
    {
     for(int i=1;i<=n;i++)
     {
      In[i]=salary[i]=0;
      if(!G[i].empty())
        G[i].clear();
     }
     for(int i=1;i<=m;i++)
      {
      int a,b;
      scanf("%d%d",&a,&b);
      G[b].push_back(a);
      In[a]++;
      } 
     Topsort();                           
    }
return 0;
}