cf472C Design Tutorial: Make It Nondeterministic

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 thei-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

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.

给出n个人，每个人有两个字符串的人名可以用，再给出一个顺序，要求按照这个顺序取人名，每人取一个，使得字符串排列严格递增

就是sb题！但是我看不懂英文蛋疼了好久……唉文化课不行伤不起

我的代码好长……在可行的情况下一定是取字符串小的好，每次比较两个字符串和当前的解的大小，然后取一个小的。如果两个都不行，直接打‘NO’

#include<cstdio>
#include<iostream>
#include<cstring>
#include<cstdlib>
#include<algorithm>
#include<cmath>
#include<queue>
#include<deque>
#include<set>
#include<map>
#include<ctime>
#define LL long long
#define inf 0x7ffffff
#define pa pair<int,int>
using namespace std;
struct people{
	char ch1[51],ch2[51];
}a[100010];
LL n;
LL s[100010];
char now[51];
inline LL read()
{
    LL x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
inline int cmp(char a[51],char b[51])
{
	int l1=strlen(a+1);
	int l2=strlen(b+1);
	for (int i=1;i<=min(l1,l2);i++)
	{
		if (a[i]<b[i])return 1;
		if (a[i]>b[i])return -1;
	}
	if (l1<l2)return 1;
	if (l1>l2)return -1;
	return 0;
}
int main()
{
	n=read();
	for (int i=1;i<=n;i++)scanf("%s%s",a[i].ch1+1,a[i].ch2+1);
	for (int i=1;i<=n;i++)s[i]=read();
	now[1]='A';
	for (int i=1;i<=n;i++)
	  {
	  	int aaa=s[i];
	  	char work1[51],work2[51];
	  	memset(work1,0,sizeof(work1));
	  	memset(work2,0,sizeof(work2));
	  	if (cmp(a[aaa].ch1,a[aaa].ch2)==1)
	  	{
	  		for(int j=1;j<=strlen(a[aaa].ch1+1);j++)work1[j]=a[aaa].ch1[j];
	  		for(int j=1;j<=strlen(a[aaa].ch2+1);j++)work2[j]=a[aaa].ch2[j];
	  	}else
	  	{
		    for(int j=1;j<=strlen(a[aaa].ch2+1);j++)work1[j]=a[aaa].ch2[j];
		    for(int j=1;j<=strlen(a[aaa].ch1+1);j++)work2[j]=a[aaa].ch1[j];
	  	}
	  	int l1=strlen(work1+1);
	  	int l2=strlen(work2+1);
	  	if (cmp(now,work2)==-1)
	  	{
	  		printf("NO");
	  		return 0;
	  	}else
	  	{
	  		if (cmp(now,work1)==-1)
	  		{
	  			memset(now,0,sizeof(now));
	  			for (int j=1;j<=l2;j++)now[j]=work2[j];
	  		}else
	  		{
	  			memset(now,0,sizeof(now));
	  			for (int j=1;j<=l1;j++)now[j]=work1[j];
	  		}
	  	}
	  }
	  printf("YES");
}