CodeForces 362A - Two Semiknights Meet

A boy Petya loves chess very much. He even came up with a chess piece ofhis 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, 2squaresbackward
and 2 to the right and 2 squares
backward and 2 to the left. Naturally, the semiknight cannotmove beyond the limits of the chessboard.

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

Petya prepared multiple chess boards. Help Petya find out whether thesemiknights 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. Eachboard is described by a matrix of characters, consisting of 8 rows and 8columns. The matrix consists of characters ".", "#",
"K", representing an empty good square, a bad square and thesemiknight's position, correspondingly. It is guaranteed that matrix containsexactly 2 semiknights.
The semiknight's squares areconsidered 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 test(s)

input

2

........

........

......#.

K..##..#

.......#

...##..#

......#.

K.......

........

........

..#.....

..#..#..

..####..

...##...

........

....K#K#
output

YES

NO
Note

Consider the first board from the sample. We will assume the rows andcolumns of the matrix to be numbered 1 through 8 from top to bottom and fromleft to right, correspondingly. The knights can meet, for example, in square (2, 7).
Thesemiknight from square (4, 1) goes tosquare (2, 3)
and thesemiknight goes from square (8, 1) tosquare (6, 3).
Thenboth semiknights go to (4,5) but thissquare is bad, so they move together to square (2, 7).

On the second board the semiknights will never meet.

思路：

一次bfs，判断第一个K能否到达第二个K的位置，且步数为偶数

代码：
#define _CRT_SECURE_NO_WARNINGS

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include <queue>
using namespace std;

const int N = 10005;

char c[15][15];
bool k[15][15];
bool p[15][15];
int tep[15][15];

int mx[] = { 2, -2, -2, 2 };
int my[] = { 2, 2, -2, -2 };

struct node
{
int x, y, step;
};

bool check(int x, int y)
{
return (x >= 0 && y >= 0 && x<8 && y<8);
}

int main()
{
int t;
scanf("%d", &t);

while (t--)
{
memset(tep, -1, sizeof(tep));
int i, j;
int x[2], y[2], l = 0;
for (i = 0; i<8; i++)
scanf("%s", c[i]);

for (i = 0; i<8; i++)
{
for (j = 0; j<8; j++)
if (c[i][j] == 'K')
{
x[l] = i;
y[l++] = j;
if (l == 2) break;
}
}

memset(k, true, sizeof(k));
queue<node> q;
node now, temp;
bool flag = false;
int tx, ty;
now.x = x[0];
now.y = y[0];
now.step = 0;
q.push(now);

while (!q.empty())
{
now = q.front();
q.pop();
k[now.x][now.y] = false;
tep[now.x][now.y] = now.step;
if (now.x == x[1] && now.y == y[1] && now.step % 2 == 0)
{
flag = true;
break;
}

for (i = 0; i<4; i++)
{
tx = now.x + mx[i];
ty = now.y + my[i];

if (check(tx, ty) && k[tx][ty])
{
temp.x = tx;
temp.y = ty;
temp.step = now.step + 1;
q.push(temp);
}
}
}

if (flag)
printf("YES\n");
else
printf("NO\n");
}

return 0;
}