HDU 2647 Reward【反向拓扑排序】

最新推荐文章于 2022-06-20 23:22:24 发布

转载最新推荐文章于 2022-06-20 23:22:24 发布 · 135 阅读

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原文链接：http://www.cnblogs.com/Roni-i/p/9180427.html

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#java

本文介绍了一个典型的图论问题：如何在满足一系列约束条件下，为员工分配最少的奖励金额。通过使用拓扑排序的方法解决该问题，并给出了完整的代码实现。

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Reward
Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 11836    Accepted Submission(s): 3804


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
 

Author
dandelion
 

Source
曾是惊鸿照影来

【题意】：老板打算给员工们发奖励，但是那些员工会去比较各自的奖励，并且会存在一些要求，例如，a员工（后继）的奖励要比b员工的奖励多。然而老板想要知道自己最少需要多少钱俩奖励员工。那么，老板决定每个员工的最低奖励为888元，这是个吉利的数字，现在第一行输入n跟m，n代表员工的数量，m代表有m个要求，接下来的m行就是员工的要求。每行有两个数a，b，表示a的奖励一定要比b的多。最后让你输出最低的资金。如果不能达到员工的要求及输出-1。

【分析】：给你n点m条有向边，叶子点(出度为0)上的值为888，父亲点为888+1，依次计算... 让你求最小的值，但是中间出现有向环就不行了，输出-1。拓扑排序队列实现，因为叶子是最小的，所以把边反向就可以求了。 a的工资比b高，所以输入a b ，建边b——>a 反向建图，然后top排序分层次；第一次的工资为888（最低），第二层的工资 + 1，后面一样

【代码】：

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e5 + 10;
const int mod = 142857;
int t,n,m,k,x,u,v,num,ans;
vector<int> G[maxn];
int inDeg[maxn];
int sum[maxn];
queue<int> q;

int topSort()
{
    num=0;
    while(!q.empty()) q.pop();
    for(int i=1;i<=n;i++) if(!inDeg[i]) q.push(i);
    while(!q.empty())
    {
        num++;
        int now = q.front();
        q.pop();
        for(int i=0;i<G[now].size();i++)
        {
            int nxt = G[now][i];
            if(--inDeg[nxt] == 0)
            {
                q.push(nxt);
                sum[nxt]=sum[now]+1;
            }
        }
    }
    if(num != n) return -1;
    int ans=0;
    for(int i=1;i<=n;i++)
        ans+=sum[i]+888;
    return ans;
}

int main()
{
    while(~scanf("%d%d",&n,&m))
    {
        memset(inDeg,0,sizeof(inDeg));
        memset(sum,0,sizeof(sum));
        for(int i=1;i<=n;i++) G[i].clear();
        while(m--)
        {
            scanf("%d%d",&u,&v);
            G[v].push_back(u); //反向建边
            inDeg[u]++;
        }
        printf("%d\n",topSort());
    }
}

转载于:https://www.cnblogs.com/Roni-i/p/9180427.html