2 seconds
256 megabytes
standard input
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: 
.
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).
If it is possible, output "YES", otherwise output "NO".
3 gennady korotkevich petr mitrichev gaoyuan chen 1 2 3
NO
3 gennady korotkevich petr mitrichev gaoyuan chen 3 1 2
YES
2 galileo galilei nicolaus copernicus 2 1
YES
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
NO
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
YES
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.
题解
字符串模拟，话说字符串函数很好用的说。
唉！话说那个……题意开始没理解，白扔了很多分。
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<iostream>
#include<cmath>
#include<algorithm>
using namespace std;
int n,p[100002];
char a[100002][2][55],c[55];
void init()
{
scanf("%d",&n);
int i;
for(i=1;i<=n;i++)
{scanf("%s%s",a[i][0],a[i][1]);
if(strcmp(a[i][0],a[i][1])==1)
{strcpy(c,a[i][0]); strcpy(a[i][0],a[i][1]); strcpy(a[i][1],c);}
}
//for(i=1;i<=n;i++) {puts(a[i][0]); puts(a[i][1]);}
for(i=1;i<=n;i++) scanf("%d",&p[i]);
}
void work()
{
int i,x=p[1],y=0;
for(i=2;i<=n;i++)
{if(strcmp(a[p[i]][1],a[x][y])==-1) {printf("NO\n"); return ;}
if(strcmp(a[p[i]][0],a[x][y])==1) {x=p[i]; y=0;}
else {x=p[i]; y=1;}
}
printf("YES\n");
}
int main()
{
init(); work();
return 0;
}