Reward HDU杭电2647【反向拓扑+队列】

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

//我的思路是：先反向拓扑，然后一次一次的把入度为零的找出来，加上888，第二次找到的就加889，依此类推

#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;

struct Reward
{
	int next;
	int follow;
}arr[20200];
int head[20200];
int indegree[11000];

int st;
int n;
int sum;
bool topo_sort()
{
	int j=1,k;
	int temp;
	queue<int>q;
	while(j<=n)
	{
		temp=-1;
		for(int i=1;i<=n;++i)
		{
			if(indegree[i]==0)
			{
				sum+=st;
				temp=i;
				q.push(i);
				indegree[i]=-1;
				++j;
			}
		}
		st++;
		if(temp==-1) return 0;
		while(!q.empty())
		{
			int pos=q.front();
			q.pop();
			for(k=head[pos];k!=-1;k=arr[k].next)
			{
				int v=arr[k].follow;
				indegree[v]--;
			}	
		}
	}
	return 1;
}
int main()
{
	int a,b,m;
	while(~scanf("%d%d",&n,&m))
	{
		st=888;sum=0;
		memset(head,-1,sizeof(head));
		memset(indegree,0,sizeof(indegree));
		for(int i=1;i<=m;++i)
		{
			scanf("%d%d",&a,&b);
			arr[i].next=head[b];
			arr[i].follow=a;//反向存边
			head[b]=i;
			indegree[a]++;
		}
		if(!topo_sort()) printf("-1\n");
		else printf("%d\n",sum);
	}
	return 0;
}