hdu 2647 Reward 逆拓扑排序+队列

最新推荐文章于 2021-08-28 13:42:41 发布

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本文介绍了一个奖金分配算法案例，解决老板如何在满足员工奖金比较需求的同时，使用最少的资金完成分配的问题。该问题通过构建有向图并利用拓扑排序进行解决。

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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.
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

老板要给很多员工发奖金，但是部分员工有个虚伪心态，认为自己的奖金必须比某些人高才心理平衡；但是老板很人道，想满足所有人的要求，并且很吝啬，想画的钱最少
输入若干个关系
a b
a c
c b
意味着a 的工资必须比b的工资高同时a 的工资比c高； c的工资比b高
不可行的时候输出 -1

如果队列能弹出的不够n那么就是-1，还有就是这里的拓扑应该是按照队列进队的，如果是用的优先队列，后进的反而先出，那么金额数会被影响一个点连着两个888，如果那个点先出去把以前的888改掉了。。那就gg了，所以用队列就先，反正没有编号的要求

#include <bits/stdc++.h>

using namespace std;

vector<int> v[10010];
queue <int> Q;
int in[10010];
int mon[10010];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(in,0,sizeof(in));
        for(int i=1;i<=n;i++)
            v[i].clear();
        while(!Q.empty()) Q.pop();
        for(int i=1;i<=m;i++)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            v[b].push_back(a);
            in[a]++;
        }
        for(int i=1;i<=n;i++)
            if(!in[i]) Q.push(i),mon[i]=888;
        int tot=0;
        int sum=0;
        while(!Q.empty())
        {
            int top=Q.front();
            sum+=mon[top];
            tot++;
            Q.pop();
            for(int i=0;i<v[top].size();i++)
            {
                in[v[top][i]]--;
                mon[v[top][i]]=mon[top]+1;
                if(!in[v[top][i]])
                    Q.push(v[top][i]);
            }
        }
        if(tot<n)
            printf("-1\n");
        else
        {
            printf("%d\n",sum );
        }
    }  
}