HDU2647 topsort

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

拓扑排序专题

vector 存图 注意vector的使用

算是拓扑排序的模板题目吧

多线程的  此题精妙在L数组 L初始为0

#include<bits/stdc++.h>
using namespace std;
vector<int> mp[20005];
int n,m;
int a,b;
int in[20005];
int L[20005];
int flag;
int re;
struct node
{
    int sum;
};
queue<node>q;
struct node N,now;
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(int i=1;i<=n;i++)
            mp[i].clear();
        memset(in,0,sizeof(in));
        memset(L,0,sizeof(L));
        for(int i=0;i<m;i++)
        {
            scanf("%d%d",&a,&b);
            mp[b].push_back(a);
            in[a]++;
        }
        for(int i=1;i<=n;i++)
        {
            if(in[i]==0)
            {
                N.sum=i;
                q.push(N);
            }
        }
        int jishu=n;
        int temp;
           while(!q.empty())
            {
                now=q.front();
                q.pop();
                jishu--;
                temp=L[now.sum];
                //cout<<now.sum<<endl;
                for(unsigned int i=0;i<mp[now.sum].size();i++)
                {
                     if(--in[mp[now.sum][i]]==0)
                      {
                          N.sum=mp[now.sum][i];
                          L[N.sum]=temp+1;
                          q.push(N);
                      }
                }
            }
            re=0;
            if(jishu>0)
                printf("-1\n");
            else
            {
                for(int i=1;i<=n;i++)
                    re+=L[i];
                printf("%d\n",re+888*n);
            }
}
    return 0;
}