# hdu2647（链式前向星+拓扑排序）Reward

2018年12月23日 1点热度 0条评论 来源: abns

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

``````#include<iostream>
#include<string.h>
#define maxn 200005
using namespace std;
int info[maxn/2];
struct edge{
int to;
int next;
int w;
}eg[maxn];
int d;
eg[d].to=u;
}//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]--;
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(eg,0,sizeof(eg));
memset(info,0,sizeof(info));
int a,b;
for(int i=1;i<=m;i++){
scanf("%d%d",&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;
}``````

原文作者：abns
原文地址: https://blog.csdn.net/qq_41453511/article/details/85224330
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