杭电 2647 Reward （拓扑排序反着排）

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

大意：
　　农场主要发工资，但是有一定规则，有些人得工资要比一些人高，每个人工资最低是888.先输入两个整数n,m,代表工人的数量编号1~n和m种规则，接下来的m行每行有两个整数a，b表示a的工资比b高，输入最小钱数，若不存在输出-1。

因为无法确定工资最高的人工资是多少，所以反着排序，让工资最小的在前面，输入a，b时记录b约束a，前驱为0的人工资为888，依次累加。

#include<cstdio>
#include<queue>
#include<string.h>
using namespace std;
int n,m,i,j,num[20010],head[20010],mon[20010],sum,ans;
struct stu
{
    int to,next;
}st[20010];
void init()
{
    memset(num,0,sizeof(num));
    memset(head,-1,sizeof(head));
    memset(mon,0,sizeof(mon));
    sum=0;
    ans=0;
}
void add(int a,int b)
{
    st[i].to=b;
    st[i].next=head[a];
    head[a]=i;
    num[b]++;
}
void topo()
{
    int i,j;
    queue<int>que;
    while(!que.empty())
    {
        que.pop();
    }
    for(i = 1 ; i <= n ; i++)
    {
        if(num[i] == 0)
        {
            mon[i]=888;
            que.push(i);
        }
    }
    while(!que.empty())
    {
        ans++;
        m=que.front();
        que.pop();
        sum+=mon[m];
        for(i = head[m] ; i != -1 ; i = st[i].next)
        {
            if(--num[st[i].to] == 0)
            {
                que.push(st[i].to);
                mon[st[i].to]=mon[m]+1;
            }
        }
    }
    if(ans == n)
        printf("%d
",sum);
    else
        printf("-1
");
}
int main()
{
    int a,b;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        init();
        for(i = 0 ; i < m ; i++)
        {
            scanf("%d %d",&a,&b);
            add(b,a);
        }
        topo();
    }
    
}