Topological Sorting

最新推荐文章于 2025-02-11 14:17:49 发布

原创最新推荐文章于 2025-02-11 14:17:49 发布 · 143 阅读

0 ·

CC 4.0 BY-SA版权

日场比赛专栏收录该内容

129 篇文章

订阅专栏

本文介绍了一种验证学习路径正确性的方法，通过检查一系列学科之间的依赖关系，判断给出的学习顺序是否符合预设条件。该方法适用于编程竞赛训练等场景。

Michael wants to win the world championship in programming and decided to study N subjects (for convenience we will number these subjects from 1 to N). Michael has worked out a study plan for this purpose. But it turned out that certain subjects may be studied only after others. So, Michael’s coach analyzed all subjects and prepared a list of M limitations in the form “ s_i u_i” (1 ≤ s_i, u_i ≤ N; i = 1, 2, …, M) , which means that subject s_i must be studied before subject u_i.

Your task is to verify if the order of subjects being studied is correct.

Remark. It may appear that it’s impossible to find the correct order of subjects within the given limitations. In this case any subject order worked out by Michael is incorrect.

Limitations
1 ≤ N ≤ 1000; 0 ≤ M ≤ 100000.

Input

The first line contains two integers N and M ( N is the number of the subjects, M is the number of the limitations). The next M lines contain pairs s_i, u_i, which describe the order of subjects: subject s_i must be studied before u_i. Further there is a sequence of N unique numbers ranging from 1 to N — the proposed study plan.

Output

Output a single word “YES” or “NO”. “YES” means that the proposed order is correct and has no contradictions with the given limitations. “NO” means that the order is incorrect.

Example

input	output
5 6 1 3 1 4 3 5 5 2 4 2 1 2 1 3 4 5 2	YES
5 6 1 3 1 4 3 5 5 2 4 2 1 2 1 2 4 5 3	NO

AC代码（记下每一个数字的位置对比就ok了）

#include <iostream>
#include <bits/stdc++.h>
using namespace std;

int main()
{
    int s[100000+10], u[100000+10], i,  a[1000+10], x;
    int n, m, f;
    while(cin>>n>>m)
    {
      f = 1;
     for(i = 1;i<=m;i++)
     {
      cin>>s[i]>>u[i];
     }
     for(i = 1;i<=n;i++)
     {
      cin>>x;
      a[x] = i;
     }
     for(i = 1;i<=m;i++)
     {
      if(a[s[i]]>a[u[i]])
      {
       f = 0;
       break;
      }
     }
     if(f==0)
     printf("NO\n");
     else
     printf("YES\n");
    }
    return 0;
}