HDOJ 题目2647 Reward（拓扑排序，邻接表建图）

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本博客讨论了工厂老板如何在春节来临之际，通过最小化成本公平地为员工分配奖励，同时满足员工间的奖励比较需求。通过输入员工数量、奖励需求数量及具体的奖励比较条件，读者可以了解到如何计算最小成本的分配方案。

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Reward

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4339 Accepted Submission(s): 1329

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

Author

dandelion

Source

曾是惊鸿照影来

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ac代码

#include<stdio.h>
#include<string.h>
#include<queue>
#include<string>
#include<iostream>
using namespace std;
struct s
{
	int to,next;
}e[30000];
int dig[10005],head[10005],a[10005],cot,n,m;
void add(int a,int b)
{
	e[cot].to=b;
	e[cot].next=head[a];
	head[a]=cot++;
}
int main()
{
	//int n,m;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		int i,j,k,v,sum,ans=0;
		queue<int>q;
		memset(a,0,sizeof(a));
		memset(head,-1,sizeof(head));
		memset(dig,0,sizeof(dig));
		cot=0;
		sum=888*n;
		for(i=0;i<m;i++)
		{
			int u,v;
			scanf("%d%d",&v,&u);
			add(u,v);
			dig[v]++;
		}
		for(i=1;i<=n;i++)
		{
			if(dig[i]==0)
				q.push(i);
		}
		while(!q.empty())
		{
			v=q.front();
			sum+=a[v];
			q.pop();
			ans++;
			for(i=head[v];i!=-1;i=e[i].next)
			{
				dig[e[i].to]--;
				if(dig[e[i].to]==0)
				{
					q.push(e[i].to);
					a[e[i].to]=a[v]+1;
				}
			}
		}
		if(ans!=n)
			sum=-1;
		printf("%d\n",sum);
	}
}