http://acm.hdu.edu.cn/showproblem.php?pid=2647

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

题目大致意思是有一个老板要给n个人发钱，下面输入m行 每行的前一个要比后一个发的多；

这样就可以形成一个有向图，钱最少的那个人无入度，用拓扑排序找到点，然后主要是想练习一下链式前向星

#include<iostream>
#include<string.h>
#define maxn 200005
using namespace std;
int info[maxn/2];
int head[maxn/2];
struct edge{
	int to;
	int next;
	int w;
}eg[maxn];
int d;
void add(int u,int v){
	eg[d].to=u;
	eg[d].next=head[v];
	head[v]=d++;
}//链式前向星
int n;
int Ts(){
	for(int i=1;i<=n;i++){
		int k=1;
		while(info[k]) k++;//找到入度为0的点
		if(k>n) return -1;
		info[k]--;
		for(int j=head[k];j!=-1;j=eg[j].next){
			int tmp=eg[j].to;
			info[tmp]--;
			eg[tmp].w=max(eg[tmp].w,eg[k].w+1);//找到最大的值
		}
	}
	return 0;
}
int main(){
	int m;
	while(~scanf("%d%d",&n,&m)){
	memset(head,-1,sizeof(head));
	memset(eg,0,sizeof(eg));
	memset(info,0,sizeof(info));
	int a,b;
	for(int i=1;i<=m;i++){
		scanf("%d%d",&a,&b);
		add(a,b);
		info[a]++;
	}
	int ans=Ts();
	if(ans==-1){
		printf("-1\n"); 
	}
	else{
		int sum=0;
		for(int i=1;i<=n;i++){
			sum+=eg[i].w;
		}
		printf("%d\n",sum+888*n);
	}
	}
	return 0;
}