CodeForces 508A Pasha and Pixels

Pasha and Pixels
Time Limit:2000MS Memory Limit:262144KB 64bit IO Format:%I64d
& %I64u
Submit Status Practice CodeForces
508A

Description

Pasha loves his phone and also putting his hair up... But the hair is now irrelevant.

Pasha has installed a new game to his phone. The goal of the game is following. There is a rectangular field consisting of n row with mpixels
in each row. Initially, all the pixels are colored white. In one move, Pasha can choose any pixel and color it black. In particular, he can choose the pixel that is already black, then after the boy's move the pixel does not change, that is, it remains black.
Pasha loses the game when a 2 × 2 square consisting of black pixels is formed.

Pasha has made a plan of k moves, according to which he will paint pixels. Each turn in his plan is represented as a pair of numbers i and j,
denoting respectively the row and the column of the pixel to be colored on the current move.

Determine whether Pasha loses if he acts in accordance with his plan, and if he does, on what move the 2 × 2 square consisting of black pixels is formed.

Input

The first line of the input contains three integers n, m, k (1 ≤ n, m ≤ 1000, 1 ≤ k ≤ 105) —
the number of rows, the number of columns and the number of moves that Pasha is going to perform.

The next k lines contain Pasha's moves in the order he makes them. Each line contains two integers i and j (1 ≤ i ≤ n, 1 ≤ j ≤ m),
representing the row number and column number of the pixel that was painted during a move.

Output

If Pasha loses, print the number of the move when the 2 × 2 square consisting of black pixels is formed.

If Pasha doesn't lose, that is, no 2 × 2 square consisting of black pixels is formed during the given k moves, print 0.

Sample Input

Input

2 2 4
1 1
1 2
2 1
2 2

Output

4

Input

2 3 6
2 3
2 2
1 3
2 2
1 2
1 1

Output

5

Input

5 3 7
2 3
1 2
1 1
4 1
3 1
5 3
3 2

Output

0

题意：一个矩阵，刚开始输入它的行和列， 都是白色，一个数k, 来涂色， 接下来是k个输入，每次一个i, j; 代表位置， 将i, j 的位置涂黑， 涂过黑色就不变了，只要出现2×2 被涂黑过，就输出到现在的步数；当然k个输入都要输入完毕才可以，如果k个输入都输入完了，也没出现2 *2 的矩阵是黑色，就输出0；

#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <string.h>
#include <queue>
#include <stack>
using namespace std;
bool num[1010][1010];
int n, m, k;

bool exist(int x, int y){
if(x - 1 > 0 && y - 1 > 0 && num[x -1][y -1] && num[x - 1][y] && num[x][y - 1])	return true;
else if(x -1 > 0 && y + 1 <= m && num[x -1][y] && num[x - 1][y + 1] && num[x][y + 1]) return true;
else if(x + 1 <= n && y - 1 > 0 && num[x][y - 1] && num[x + 1][y - 1] & num[x + 1][y])	return true;
else if(x + 1 <= n && y + 1 <= m && num[x][y + 1] && num[x + 1][y + 1] && num[x + 1][y])	return true;
else
return false;
}
int main(){

while(~scanf("%d%d%d", &n, &m, &k)){

memset(num, false, sizeof(num));
int i, j;
int step = 0;
int flag = 0;
while(k--){
cin >> i >> j;
num[i][j] = true;
if(flag == 0){
++step;
if(exist(i, j)){
flag = 1;
}
}
}
if(flag == 0){
cout << "0" << endl;
}
else
cout << step << endl;
}
return 0;
}