1280. Topological Sorting

Time limit: 1.0 second Memory limit: 64 MB

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.

Samples

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

Problem Author: © Sergey G. Volchenkov, 2003(volchenkov@yandex.ru); Vladimir N. Pinaev, 2003(vpinaev@mail.ru; http://www.pic200x.chat.ru); Michael Y. Kopachev, 2003 (mkopachev@krista.ru). Problem Source: 2003-2004 ACM Central Region of Russia Quarterfinal Programming Contest, Rybinsk, October 15-16, 2003

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拓扑排序（用链表）

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 1 #include<iostream>
 2 #include<string>
 3 #include<cstring>
 4 #include<cmath>
 5 #include<cctype>
 6 #include<cstdio>
 7 #include<stack>
 8 #include<queue>
 9 #include<vector>
10 using namespace std;
11 const int maxn=10005;
12 int ind[maxn];
13 struct node
14 {
15     int ve;
16     struct node *next;
17     node()
18     {
19         ve=0;
20         next=NULL;
21     }
22 };
23 node *edge[maxn];
24 void insert(int x,int y)//链表的链接
25  {
26      node *temp;
27      temp=new node(); 
28      temp->ve=y;
29      temp->next=edge[x];
30      edge[x]=temp;
31      ind[y]++;
32 
33  }
34 int main()
35 {
36     int i,j,be,en,n,m,s;
37     memset(ind,0,sizeof(ind));
38     cin>>n>>m;
39     for(i=1;i<=m;i++)
40      {
41          cin>>be>>en;
42          insert(be,en);
43      }
44     for(i=1;i<=n;i++)
45      {
46          cin>>s;
47          if(ind[s]==0)
48           {
49               while(edge[s])
50                {
51                    ind[edge[s]->ve]--;
52                    edge[s]=edge[s]->next;
53                }
54           }
55          else
56            {
57                 cout<<"NO"<<endl;
58                 return 0;
59            }
60      }
61      cout<<"YES"<<endl;
62      return 0;
63 }

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