ZCMU-1342-Two Semiknights Meet

1342: Two Semiknights Meet

Time Limit: 1 Sec  Memory Limit: 128 MB
Submit: 56  Solved: 17

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Description

A boy Petya loves chess very much. He even came up with a chess piece of his own, a semiknight. The semiknight can move in any of these four directions: 2 squares
forward and 2 squares to the right, 2 squares
forward and 2 squares to the left, 2 squares
backward and 2 to the right and 2 squares
backward and 2 to the left. Naturally, the semiknight cannot move beyond the limits of the chessboard.

Petya put two semiknights on a standard chessboard. Petya simultaneously moves with both semiknights. The squares are rather large, so after some move the semiknights can meet, that is, they can end up in the same square.
After the meeting the semiknights can move on, so it is possible that they meet again. Petya wonders if there is such sequence of moves when the semiknights meet. Petya considers some squares bad. That is, they do not suit for the meeting. The semiknights
can move through these squares but their meetings in these squares don't count.

Petya prepared multiple chess boards. Help Petya find out whether the semiknights can meet on some good square for each board.

Please see the test case analysis.

Input

The first line contains number t (1 ≤ t ≤ 50)
— the number of boards. Each board is described by a matrix of characters, consisting of 8 rows and 8 columns. The matri
d517
x consists of characters ".", "#",
"K", representing an empty good square, a bad square and the semiknight's position, correspondingly. It is guaranteed that matrix contains exactly 2 semiknights.
The semiknight's squares are considered good for the meeting. The tests are separated by empty line.

Output

For each test, print on a single line the answer to the problem: "YES", if the semiknights can meet and "NO"
otherwise.

Sample Input

2
........

........

......#.

K..##..#

.......#

...##..#

......#.

K.......

........

........

..#.....

..#..#..

..####..

...##...

........

....K#K#

Sample Output

YES
NO

【解析】
本题就是叫你去判断两个人能不能见面，这两个人分别用K来表示，一个人在走的同时另一个人也在走，他们的走法是田字走法，斜着走问你他们两能不能到达同一个地点。这个时候我们总会有点感觉斜着走两格相当于x轴走了两个y轴走了两格。所以如果他们的横坐标和纵坐标都是4的倍数那肯定是可以见得到的，否则肯定是不可以的。有可能会有人奇怪为什么我输入字符又用了C++的输入，那是因为C++输入字符不会读入换行符类似的题目详见我另一篇文章http://blog.csdn.net/ZCMUCZX/article/details/53997118?locationNum=1&fps=1

#include<iostream>
#include<string>
#include<cstdio>
#include<cstring>
#include<cmath>
using namespace std;
int main()
{
int t,i,j,x1,y1,x2,y2,flag,p,q;
char c;
scanf("%d",&t);
while(t--)
{
flag=0;
for(i=0;i<8;i++)
{
for(j=0;j<8;j++)
{
cin>>c;
if(c=='K'&&flag==0)
{
flag=1;
x1=i;
y1=j;
}
else if(c=='K'&&flag==1)
{
x2=i;
y2=j;
}
}
}
p=fabs(x1-x2);
q=fabs(y1-y2);
if(p%4==0&&q%4==0)
{
printf("YES\n");
}
else
{
printf("NO\n");
}
}
return 0;
}