Codeforces Round #270 1003

Codeforces Round #270 1003

C. Design Tutorial: Make It Nondeterministic

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

A way to make a new task is to make it nondeterministic or probabilistic. For example, the hard task of Topcoder SRM 595, Constellation, is the probabilistic version of a convex hull.

Let's try to make a new task. Firstly we will use the following task. There are n people, sort them by their name. It is just an ordinary sorting problem, but we can make it more interesting by adding nondeterministic element. There are n people, each person will use either his/her first name or last name as a handle. Can the lexicographical order of the handles be exactly equal to the given permutation p?

More formally, if we denote the handle of the i-th person as hi, then the following condition must hold:

.

Input

The first line contains an integer n(1 ≤ n ≤ 105) — the number of people.

The next n lines each contains two strings. The i-th line contains strings fi and si(1 ≤ |fi|, |si| ≤ 50) — the first name and last name of the i-th person. Each string consists only of lowercase English letters. All of the given 2n strings will be distinct.

The next line contains n distinct integers: p1, p2, ..., pn(1 ≤ pi ≤ n).

Output

If it is possible, output "YES", otherwise output "NO".

Sample test(s)

Input

3 
gennady korotkevich 
petr mitrichev 
gaoyuan chen 
1 2 3

Output

NO

Input

3 
gennady korotkevich 
petr mitrichev 
gaoyuan chen 
3 1 2

Output

YES

 

pair的用法，sort按first排序

名姓分别存，排序后从前往后扫存在可行能扫到底每个人都被扫到。

 #include<iostream>
 #include<cstdio>
 #include<cstring>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 int n;
 pair<string,int> mp[];

 int main()
 {
     scanf("%d",&n);
     for(int i = ;i<=(n<<);i++)
     {
         cin>>mp[i].first;
         mp[i].second = (i+)/;
     }
     sort(mp+,mp+*n+);
     int rk;
     int j = ;
     int ct = ;
     for(int i = ;i<=n;i++)
     {
         scanf("%d",&rk);
         for(;j<=*n;j++)
         {
             if(mp[j].second==rk)
                {
                    ct++;
                    break;
                }
         }
     }
     if(ct==n) puts("YES");
     else puts("NO");
     return ;
 }