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.
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;
}