hdu 2647 Reward (topsort)

Reward
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 13132    Accepted Submission(s): 4199

Problem 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

 

C\C++:

#include <map>
#include <queue>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstring>
#include <climits>
#include <iostream>
#include <algorithm>
#define INF 0xffffff
using namespace std;

const int my_max = 20010;
int n, m, a, b, my_cnt, my_indeg[my_max];
vector <int> my_G[my_max];

int topsort()
{
    my_cnt = 0;
    int my_ans = 0, t = 887;
    queue <int> Q;
    while (t ++)
    {
        int flag = -1, my_temp = 0;
        for (int i = 1; i <= n; ++ i)
            if (my_indeg[i] == 0)
            {
                my_indeg[i] = -1;
                flag = 1;
                Q.push(i);
                my_cnt ++;
                my_temp ++;
            }

        if (flag == -1) break;
        my_ans += my_temp * t;

        while (my_temp --)
        {
            int my_now = Q.front();
            for (int i = 0; i < my_G[my_now].size(); ++ i)
                my_indeg[my_G[my_now][i]] --;
            my_G[my_now].clear();
            Q.pop();
        }
    }
    if (my_cnt == n)
        return my_ans;

    for (int i = 1; i <= n; ++ i)
        my_G[i].clear();
    return -1;
}

int main()
{
    while (~scanf("%d%d", &n, &m))
    {
        memset(my_indeg, 0, sizeof(my_indeg));
        while (m --)
        {
            scanf("%d%d", &a, &b);
            ++ my_indeg[a];
            my_G[b].push_back(a);
        }

        printf("%d\n", topsort());
    }
    return 0;
}