Hdoj 2467 Reward(拓扑排序)

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本文讨论了工厂老板如何在春节来临之际，通过使用邻接表和拓扑排序来公平且经济地分配奖励给工人，确保多劳多得的原则得以实现。详细解释了在特定条件下如何避免形成环状依赖，从而找到最低成本的解决方案。

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

Sample Output

1777
-1

题意：
一：不能在用简单的拓扑了，因为此时显然超内存。（此题内存只能开8000000的数组，不能开到一千万）只能用到邻接表。
二：注意多劳多得，可能前面的一个人后面跟着好几个，即，后面那几个人的工资是一样的
最后题意简单，就是叫你判读是否有环。

///@zhangxiaoyu
///2015/8/10

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<stack>
#include<queue>
#include<cmath>
#include<iostream>
using namespace std;
#define INF 0x7f7f7f
#define maxn 1000005

int in[10010],x[10010],money[10010];
int len,num;
int n,m;
vector<int>E[10010];

void init()
{
    for(int i=1;i<=n;i++)
    {
        in[i]=0;
        E[i].clear();
        money[i]=888;
    }
    len=0;
    num=0;
}

void topsort()///vector版本的拓扑排序模板
{
    int j;
    priority_queue<int,vector<int>,greater<int> >qq;///最小堆优队
    for(int i=1;i<=n;i++)
    {
        if(in[i]==0)
            qq.push(i);
    }
    while(!qq.empty())
    {
        j=qq.top();
        qq.pop();
        num++;
        x[++len]=j;
        for(int i=0;i<E[j].size();i++)
        {
            int k=E[j][i];
            in[k]--;
            if(money[j]+1>money[k])
                money[k]=money[j]+1;///求最少的钱，又后边的奖金比前面大，故贪心的加1就是最少的
            if(in[k]==0)
                qq.push(k);
        }
    }
}

int main()
{
    int x,y;
    while(~scanf("%d%d",&n,&m))
    {
        init();
        for(int i=1;i<=m;i++)
        {
            scanf("%d%d",&x,&y);
            E[y].push_back(x);
            in[x]++;
        }
        topsort();
        if(num<n)
            printf("-1\n");
        else
        {
            int ans=0;
            for(int i=1;i<=n;i++)
            {
                ans+=money[i];
            }
            printf("%d\n",ans);
        }
    }
    return 0;
}