hdu 2467 拓扑排序之反向建图

最新推荐文章于 2024-08-06 11:44:00 发布

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最新推荐文章于 2024-08-06 11:44:00 发布

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分类专栏：图论拓扑排序

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本文介绍了一道关于如何在满足特定条件的情况下，用最少的钱为工人分配奖励的问题。该问题通过拓扑排序解决，并实现了反向建图以确保A的奖励总是高于B。代码示例使用C++实现。

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http://acm.hdu.edu.cn/showproblem.php?pid=2647

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

Sample Output

  
   1777
-1

一道拓扑排序的裸题，解题过程中要注意a比b拿的奖励要多，要用反向建图。这里容易出错别的直接套模板就哦了。

#include <string.h>
#include <stdio.h>
#include <iostream>
using namespace std;
struct note
{
    int to;
    int next;
};
note edge[20005];
int head[20005],ip,iq;
int indegree[20005],queue[20005],money[20005];
void add(int u,int v)
{
    edge[ip].to=v,edge[ip].next=head[u],head[u]=ip++;
}
int main()
{
    int m,n,a,b;
    while(cin >>n>>m)
    {
         memset(head,-1,sizeof(head));
         ip=iq=0;
         memset(indegree,0,sizeof(indegree));
         for(int i=0;i<=n;i++)
               money[i]=888;
         for(int i=0;i<m;i++)
         {
             cin>>a >>b;
             add(b,a);
             indegree[a]++;
         }
        for(int i=1;i<=n;i++)
            if(indegree[i]==0)
                queue[iq++]=i;
        for(int i=0;i<iq;i++)
        {
            for(int k=head[queue[i]];k!=-1;k=edge[k].next)
            {
                if(--indegree[edge[k].to]==0)
                {
                    money[edge[k].to]=money[queue[i]]+1;
                    queue[iq++]=edge[k].to;
                }
            }
        }
        int sum=0;
        if(iq ==n)
        {
            for(int i=1;i<=n;i++)
            {
                sum+=money[i];
            }
            printf("%d\n",sum);
       }
        else
            printf("-1\n");
    }
    return 0;
}