【題意】：

就是有一個老闆，給自己的工人發獎金，起步888，而且，a要比b領得多，問你怎麼發才能發最少的錢（自己不虧的 jian 商）；

【思考】：

可以建個圖，用拓撲，找到它們的入度，加上888就是那個最多的是最小的錢；這裡有兩種建圖，重要思想還是拓撲。。。

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

ac程式碼一（vector版）：

#include<cstring>
#include<cstdio>
#include<vector>

using namespace std;

const int maxn=2e4+10;

vector<int>v[maxn];
int money[maxn],vis[maxn],ans;
int n,m;

void topu_sort()
{
	ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(!vis[j]){
				vis[j]--;
				ans++;
				for(int k=0;k<v[j].size();k++)
				{
					vis[v[j][k]]--;
					if(money[v[j][k]]-money[j]<1)
						money[v[j][k]]=money[j]+1;;	
				}
			}
		}
	}
	if(ans!=n)	printf("-1\n");
	else{
		ans=0;
		for(int i=1;i<=n;i++)
			ans+=888+money[i];
		printf("%d\n",ans);
	}
}

int main()
{
	while(scanf("%d%d",&n,&m)!=EOF){
		memset(money,0,sizeof(money));
		memset(vis,0,sizeof(vis));
		for(int i=1;i<=m;i++)
		{
			int a,b;
			scanf("%d%d",&b,&a);
			v[a].push_back(b);
			vis[b]++;
		}
		topu_sort();
		for(int i=0;i<=n;i++)
			v[i].clear();
	}
	
	return 0;
}
 

ac程式碼二（鏈式前向星版）：

寫這個程式碼的時候，建錯圖了，搞得懵懵的，多虧了我家的小仙女，唉，鄙人還是太菜了，多練練

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


int head[20008],money[10008],vis[10008];
int n,m,ans;

struct node{
	int now,next;
}edge[20008];


int cnt;
void add(int u,int v){
	edge[cnt].now =v;
	edge[cnt].next =head[u];
	head[u]=cnt++;
}

void topu_sort()
{
	ans=0;
	for(int i=1;i<=n;i++){
		for(int j=1;j<=n;j++){
			if(!vis[j])
			{
				vis[j]--;
				ans++;
				for(int k=head[j];k!=0;k=edge[k].next)
					{
						vis[edge[k].now]--;
						if(money[edge[k].now]-money[j]<1)
							money[edge[k].now ]=money[j]+1;
					}
			}
		} 
	}
	if(ans!=n)	printf("-1\n");
	else{
		ans=0;
		for(int i=1;i<=n;i++)
			ans+=888+money[i];
		printf("%d\n",ans);
	}
}

int main(){
	while(scanf("%d%d",&n,&m)!=EOF){
		memset(money,0,sizeof(money));
		memset(vis,0,sizeof(vis));
		memset(head,0,sizeof(head));
		cnt=1;
		while(m--){
			int a,b;
			scanf("%d%d",&a,&b);
			add(b,a);
			vis[a]++;
		}
		topu_sort();
	}

	return 0;
}