Reward HDU

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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. 

InputOne 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.OutputFor 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


题意：输入两个数n,m。n表示公司人数，m表示不同人之间的关系。a b表示a的工资大于b.
题解：拓扑排序，找入度为0的点，所有的拓扑完没有剩余，说明没有环

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<sstream>
#include<cmath>
#include<stack>
#include<map>
#include<cstdlib>
#include<vector>
#include<string>
#include<queue>
using namespace std;

#define ll long long
#define llu unsigned long long
#define INF 0x3f3f3f3f
const double PI = acos(-1.0);
const int maxn =  1e4+10;
const int mod = 1e9+7;
int degree[maxn],add[maxn];
vector<int>G[maxn];
int n,m;
int toposort()
{
    queue<int>que;
    for(int i=1;i<=n;i++)
        if(degree[i] == 0)
            que.push(i);
    int ans = 0;
    while(que.size())
    {
        int ptr = que.front();
        que.pop();
        ans++;
        for(int i=0;i<G[ptr].size();i++)
        {
            degree[G[ptr][i]]--;
            if(degree[G[ptr][i]] == 0)
            {
                que.push(G[ptr][i]);
                add[G[ptr][i]] = add[ptr] + 1;
            }
        }
    }
    if(ans != n)
        return -1;
    int sum = 0;
    for(int i=1;i<=n;i++)
        sum += add[i] + 888;
    return sum;
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(degree,0,sizeof degree);
        memset(add,0,sizeof add);
        for(int i=1;i<=n;i++)
            G[i].clear();
        for(int i=0;i<m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            G[b].push_back(a);
            degree[a]++;
        }
        printf("%d
",toposort());
    }
}