Reward_toposort

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7610    Accepted Submission(s): 2418

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

#include<iostream>
#include<queue>
#include<string.h>
#include<stdio.h>
using namespace std;
const int N=10005;
int in[N],mo[N];
vector<int>v[N];
int cnt,sum,n;
queue<int>qu;
void init()
{
    while(!qu.empty()) qu.pop();
    for(int i=1;i<=n;i++)
    {
        v[i].clear();
        mo[i]=888;//刚开始都为888；
        in[i]=0;
    }
}
void toposort()
{
    sum=0;cnt=0;
    while(!qu.empty())
    {
        int t=qu.front();
        qu.pop();
        sum+=mo[t];
        cnt++;//没有矛盾的情况，最后应该是n
        for(int i=0;i<v[t].size();i++)
        {
            in[v[t][i]]--;
            if(in[v[t][i]]==0) qu.push(v[t][i]);
            mo[v[t][i]]=mo[t]+1;//比t多，在t的基础上加一；
        }
    }
}
int main()
{
    int m,a,b;
    while(cin>>n>>m)
    {
        init();
        for(int i=1;i<=m;i++)
        {
            cin>>a>>b;
            v[b].push_back(a);//这里应该是反的，a必须比b多。刚开始没注意，一直wa！！
            in[a]++;
        }
        int flag=0;
        for(int i=1;i<=n;i++)
        {
            if(!in[i]){qu.push(i);flag=1;}
        }
        if(flag==0) cout<<"-1"<<endl;
        else
        {
            toposort();
            if(cnt!=n) cout<<"-1"<<endl;
            else cout<<sum<<endl;
        }

    }
    return 0;
}