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
#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 ;
}