POJ3687 Labeling Balls逆拓扑排序

最新推荐文章于 2022-06-20 23:22:24 发布

nextcy

最新推荐文章于 2022-06-20 23:22:24 发布

阅读量713

点赞数

CC 4.0 BY-SA版权

分类专栏： ACM 文章标签： output each input 存储

本文链接：https://blog.youkuaiyun.com/chaiyuan414/article/details/4991097

ACM 专栏收录该内容

52 篇文章

订阅专栏

本文介绍了一种解决球体重排序问题的方法，通过逆拓扑排序算法确定每个球的正确重量标签，确保满足所有给定的重量约束条件。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

Labeling Balls

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 3961		Accepted: 910

Description

Windy has N balls of distinct weights from 1 unit to N units. Now he tries to label them with 1 to N in such a way that:

No two balls share the same label.
The labeling satisfies several constrains like "The ball labeled with a is lighter than the one labeled with b".

Can you help windy to find a solution?

Input

The first line of input is the number of test case. The first line of each test case contains two integers, N (1 ≤ N ≤ 200) and M (0 ≤ M ≤ 40,000). The next M line each contain two integers a and b indicating the ball labeled with a must be lighter than the one labeled with b. (1 ≤ a, b ≤ N) There is a blank line before each test case.

Output

For each test case output on a single line the balls' weights from label 1 to label N. If several solutions exist, you should output the one with the smallest weight for label 1, then with the smallest weight for label 2, then with the smallest weight for label 3 and so on... If no solution exists, output -1 instead.

Sample Input

Sample Output

1 2 3 4
-1
-1
2 1 3 4
1 3 2 4
这题的题意浪费了我几个小时，一直以为是输出他们的字典序列，后来才知道是输出他们的重量标号，其实就是进行

逆拓扑排序。输入的是1 3；就是说1比3轻，而在存储时记作3比1轻。然后从最后开始进行查找，程序如下：

#include<iostream>
#include<cstring>
using namespace std;
int _d[201][201],n,m,_in[201],ans[201];
int main()
{
    int ncase;
    cin>>ncase;
    while(ncase--)
    {
           memset(_d,0,sizeof(_d));
           memset(_in,0,sizeof(_in));
           cin>>n>>m;           
           int i,j,a,b,k;
           for(i=1;i<=m;i++)      //进行逆排序 
           {
                cin>>a>>b;
                if(!_d[b][a])
                {
                        _d[b][a]=1;
                        _in[a]++;
                }                        
           }
           for(i=n;i>0;i--)   //要排成n个数的序列  
           {
               for(j=n;j>0&&_in[j];j--)   //每次都对n个数进行扫描 
               {
                    if(_in[j]<0)   continue;    //如果已经被扫描出来 就往下进行 
               }
               if(j==0) break;       //没有找到符合的 
               ans[j]=i;               //标记已经找到的 
               _in[j]=-1;
               //cout<<j<<endl; 
               for(k=1;k<=n;k++)
                   if(_d[j][k])  _in[k]--;     //删除标记点的弧 
           }
           //for(i=1;i<=n;i++)  cout<<ans[i]<<" ";
           //cout<<i<<endl;
           if(i>0)  cout<<-1<<endl;
           else
           {
               for(i=1;i<=n;i++)
                  cout<<ans[i]<<" ";
               cout<<endl;
           }
    }
    return 0;
}