Problem F. Turning Grille
题目连接：
http://opentrains.snarknews.info/~ejudge/team.cgi?SID=c75360ed7f2c7022&all_runs=1&action=140
Description
‘Turning grille’ method is one of the simplest forms of transposition cipher. A variant of this method is
as follows. The sender of the message writes it onto a square sheet of paper gridded into N rows and
N columns, one character per cell. The number N is even. A grille is used for ciphering—a pierced sheet
of cardboard of the same size as the paper. There are N2/4 holes made in the grille, and one cell can be
seen through each hole. When writing a message, the grille is placed on a sheet of paper, and the letters
of the message are written into the holes from top to bottom. After all holes have been filled, the grille
is turned 90◦
clockwise, and the writing goes on. The grille can be turned three times, and the message
length does not exceed N2
.
The receiver of the message has the same grille and, performing the same manipulations, reads the
characters that appear in the holes.
A well-formed grille must not show the same paper sheet cell several times during grille rotations. However,
the holes are manufactured by hand and a master can make mistakes, especially when creating a big
grille. . .
Write a program that checks if a grille is correct.
Input
The first line of input contains the integer N, the size of the paper (4 ≤ N ≤ 1000, N is even). N lines
follow, each corresponds to a row of the grille and contains N characters . (no hole) or * (a hole). The
number of holes is N2/4.
Output
Output the line YES or NO depending on whether the grille is well-formed or not.
Sample Input
4
...
..*
....
.*..
Sample Output
YES
Hint
题意
给你一个解密卡，就是那个*就是洞洞，然后转90°，转四次，问你洞洞能不能覆盖所有点。
保证解密卡有n^2/4个洞洞。
题解：
哈哈，我说这个就是像小时候玩过的冒险小虎队的那个解密卡。
这个xjb搞一搞就好了
代码
#include <bits/stdc++.h>
using namespace std;
const int maxn = 1e3 + 15;
int N ;
char str[4][maxn][maxn];
int main(int argc, char * argv[]){
// freopen("in.txt","r",stdin);
scanf("%d",&N);
for(int i = 0 ; i < N ; ++ i){
scanf("%s" , str[0][i]);
}
for(int i = 1 ; i < 4 ; ++ i) for(int j = 0 ; j < N ; ++ j) for(int k = 0 ; k < N ; ++ k) str[i][j][k] = str[i - 1][k][N-j-1];
/* for(int i = 0 ; i < 4 ; ++ i){
for(int j = 0 ; j < N ; ++ j) cout << str[i][j] << endl;
cout << endl;
}*/
vector < pair < int , int > > vi;
for(int i = 0 ; i < 4 ; ++ i)
for(int j = 0 ; j < N ; ++ j)
for(int k = 0 ; k < N ; ++ k)
if( str[i][j][k] == '*' )
vi.push_back( make_pair( j , k ) );
sort( vi.begin() , vi.end() );
assert( vi.size() == N * N );
int C = unique( vi.begin() , vi.end() ) - vi.begin();
if( C == N * N ) printf("YES
");
else printf("NO
");
return 0;
}