CodeForces -Game on Paper

Description

One not particularly beautiful evening Valera got very bored. To amuse himself a little bit, he found the following game.

He took a checkered white square piece of paper, consisting of n × n cells. After that, he started to paint the white cells black one after the other. In total he painted m different cells on the piece of paper. Since Valera was keen on everything square, he wondered, how many moves (i.e. times the boy paints a square black) he should make till a black square with side 3 can be found on the piece of paper. But Valera does not know the answer to this question, so he asks you to help him.

Your task is to find the minimum number of moves, till the checkered piece of paper has at least one black square with side of 3. Otherwise determine that such move does not exist.

Input

The first line contains two integers n and m(1 ≤ n ≤ 1000, 1 ≤ m ≤ min(n·n, 105)) — the size of the squared piece of paper and the number of moves, correspondingly.

Then, m lines contain the description of the moves. The i-th line contains two integers xi, yi (1 ≤ xi, yi ≤ n) — the number of row and column of the square that gets painted on the i-th move.

All numbers on the lines are separated by single spaces. It is guaranteed that all moves are different. The moves are numbered starting from 1 in the order, in which they are given in the input. The columns of the squared piece of paper are numbered starting from 1, from the left to the right. The rows of the squared piece of paper are numbered starting from 1, from top to bottom.

Output

On a single line print the answer to the problem — the minimum number of the move after which the piece of paper has a black square with side 3. If no such move exists, print -1.

Sample Input

Input

4 11

1 1

1 2

1 3

2 2

2 3

1 4

2 4

3 4

3 2

3 3

4 1

Output

10

方法1：

#include <stdio.h>
#include <string.h>
#define  N 1010
int matrix

={0};
bool check(int x,int y){
//当(x,y)位于九宫格中心时，判断是否为黑九宫格
if (matrix[x-1][y-1]&&matrix[x-1][y]&&matrix[x-1][y+1]&&matrix[x][y-1]&&matrix[x][y]&&
matrix[x][y+1]&&matrix[x+1][y-1]&&matrix[x+1][y]&&matrix[x+1][y+1])
return true;
else
return false;
}
int main(){
freopen("input.txt","r",stdin);
int n,m;
int x,y,i;
bool flag;
while(scanf("%d%d",&n,&m)!=EOF){
flag=false;
memset(matrix,0,sizeof(matrix));
for (i = 1; i <=m; ++i)
{
scanf("%d%d",&x,&y);
matrix[x][y]=1;
//(x,y)可能位于九宫格的左上、上、右上、左、中、右、左下、下、右下，分别判断这
//九种情况对应的中心时候可以构成黑九宫格
if (check(x,y)||check(x+1,y+1)||check(x+1,y)||check(x+1,y-1)||check(x,y+1)
||check(x,y-1)||check(x-1,y+1)||check(x-1,y)||check(x-1,y-1))
{
if (!flag)
{
flag=true;
printf("%d\n",i);
}

}
}
if (!flag)
{
printf("-1\n");
}

}
return 0;
}

方法2：

#include <stdio.h>
#include <string.h>
#define N 1010
int matrix

={0};
//输入一个(x,y)，便将以(x,y)为中心的九宫格每个单元格＋1，
//当某个单元格达到9时，此时便形成黑九宫格。
int main()
{
freopen("input.txt","r",stdin);
int n,m;
int x,y;
bool flag;
while(scanf("%d%d",&n,&m)!=EOF){
flag = false;
memset(matrix,0,sizeof(matrix));
for (int i = 1; i <=m; ++i)
{
scanf("%d%d",&x,&y);
for (int k = x-1; k <= x+1; ++k)
{
for (int j = y-1; j <= y+1; ++j)
{
matrix[k][j]++;
//printf("i=%d,(%d,%d),matrix[%d][%d]=%d\n",i,x,y,k,j,matrix[k][j]);

if (matrix[k][j]==9)
{
if (!flag)
{
flag=true;
printf("%d\n",i);
}

}

}
}
}
if (!flag)
{
printf("-1\n");
}

}

}