Codeforces Round #270——C. Design Tutorial: Make It Nondeterministic

C. Design Tutorial: Make It Nondeterministic

time limit per test

memory limit per test

input

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 h_i, then the following condition must hold:

.

Input

The first line contains an integer n (1 ≤ n ≤ 10⁵)

The next n lines each contains two strings. The i-th line contains strings f_i and s_i (1 ≤ |f_i|, |s_i| ≤ 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

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

Output

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

Sample test(s)

Input

3gennady korotkevich petr mitrichev gaoyuan chen 1 2 3

Output

NO

Input

3gennady korotkevich petr mitrichev gaoyuan chen 3 1 2

Output

YES

Input

2galileo galilei nicolaus copernicus 2 1

Output

YES

Input

10rean schwarzer fei claussell alisa reinford eliot craig laura arseid jusis albarea machias regnitz sara valestin emma millstein gaius worzel 1 2 3 4 5 6 7 8 9 10

Output

NO

Input

10rean schwarzer fei claussell alisa reinford eliot craig laura arseid jusis albarea machias regnitz sara valestin emma millstein gaius worzel 2 4 9 6 5 7 1 3 8 10

Output

YES

Note

In example 1 and 2, we have 3 people: tourist, Petr and me (cgy4ever). You can see that whatever handle is chosen, I must be the first, then tourist and Petr must be the last.

In example 3, if Copernicus uses "copernicus" as his handle, everything will be alright.

我是贪心做的，按题目所给的顺序比较每一行的，取大的，如果顺利比完，就输出YES，否则输出NO

#include<map>  
#include<set>  
#include<list>  
#include<stack>  
#include<queue>  
#include<vector>  
#include<cmath>  
#include<cstdio>  
#include<cstring>  
#include<iostream>  
#include<algorithm>  
  
using namespace std; 

struct peo
{
	char f[55];
	char s[55];
}arr[100010];

char cur[55];

int p[100010];

int main()
{
	int n;
	while(~scanf("%d", &n))
	{
		for(int i = 1; i <= n; i++)
		{
			scanf("%s%s", arr[i].f, arr[i].s);
		}
		for(int i = 1; i <= n; i++)
		{
			scanf("%d", &p[i]);
		}
		bool flag = true;
		if(strcmp(arr[p[n]].f, arr[p[n]].s) > 0)
			strcpy(cur, arr[p[n]].f);
		else
			strcpy(cur, arr[p[n]].s);
		for(int i = n - 1; i >= 1; i--)
		{
		 	if(strcmp(cur, arr[p[i]].f) > 0 || strcmp(cur, arr[p[i]].s) > 0)
		 	{
		 		if(strcmp(cur, arr[p[i]].f) > 0 && strcmp(cur, arr[p[i]].s) > 0)
		 		{
		 			if(strcmp(arr[p[i]].f, arr[p[i]].s) > 0)
		 				strcpy(cur, arr[p[i]].f);
	 				else
	 					strcpy(cur, arr[p[i]].s);
		 		}
		 		else if(strcmp(cur, arr[p[i]].f) > 0 && strcmp(cur, arr[p[i]].s) <= 0)
		 			strcpy(cur, arr[p[i]].f);
	 			else
	 				strcpy(cur, arr[p[i]].s);
	 		}
	 		
	 		else
	 		{
		 		flag = false;
		 		break;
		 	}
		}
		if(flag)
			printf("YES\n");
		else
			printf("NO\n");
	}
	return 0;
}