CF - 472C. Design Tutorial: Make It Nondeterministic 贪心+字符串+模拟

本文链接：https://blog.youkuaiyun.com/dragon60066/article/details/59124225

本文介绍了一种非确定性排序问题，即每个人可以选择使用名或姓作为账号，并给出一个期望的字典序排列，判断该排列是否可以实现。通过贪心算法选择尽可能小的名字来构造目标序列。

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1.题目描述：

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: $\text{[math]}$ .

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".

Examples

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

input
2
galileo galilei
nicolaus copernicus
2 1

output
YES

input
10
rean 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
10
rean 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.

2.题意概述：

每个人可以选择first name 或者 last name 当自己的账号，现在给出最终的字典序排位，问能不能实现

3.解题思路：

要尽可能的最终字典序排，那么对于每个人，肯定是尽可能地选择字典序小的name，开一个string记录当前i - 1个名字，那么每次选择就贪心地选择字典序小的名字，如果字典序小的不满足再选择字典序大的名字，如果都不满足则不能实现

4.AC代码：

#include <bits/stdc++.h>
#define INF 0x3f3f3f3f
#define maxn 100100
#define N 1111
#define eps 1e-6
#define pi acos(-1.0)
#define e 2.718281828459
#define mod 32767
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
struct node
{
  int id;
  string first, last;
} p[maxn];
int main()
{
#ifndef ONLINE_JUDGE
  freopen("in.txt", "r", stdin);
  freopen("out.txt", "w", stdout);
  long _begin_time = clock();
#endif
  int n;
  while (~scanf("%d", &n))
  {
    for (int i = 1; i <= n; i++)
      cin >> p[i].first >> p[i].last;
    int tmp, flag = 1;
    scanf("%d", &tmp);
    string cur = min(p[tmp].first, p[tmp].last);
    for (int i = 0; i < n - 1; i++)
    {
      scanf("%d", &tmp);
      if (flag)
      {
        string temp = min(p[tmp].first, p[tmp].last);
        if (cur >= temp)
        {
          if (cur >= max(p[tmp].first, p[tmp].last))
            flag = 0;
          else
            cur = max(p[tmp].first, p[tmp].last);
        }
        else
          cur = temp;
      }
    }
    puts(flag ? "YES" : "NO");
  }
#ifndef ONLINE_JUDGE
  long _end_time = clock();
  printf("time = %ld ms.", _end_time - _begin_time);
#endif
  return 0;
}