Codeforces Round #288 (Div. 2) A. Pasha and Pixels

A. Pasha and Pixels

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

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 iand 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 test(s)

input

2 2 41 11 22 12 2

output

4

input

2 3 62 32 21 32 21 21 1

output

5

input

5 3 72 31 21 14 13 15 33 2

output

0

这次cf最最不能容忍的是，我的A题居然挂了，检查代码的时候发现手残啊，四个if语句让我从中间加了个else，还只加了一个

题意为给你一个n*m的矩阵，初始值为0，然后给出k个点坐标，每次把这些点的值变为1,问第几次矩阵出现个2*2的矩阵的值全为1，如果没有输出0，好了直接暴就可以了。。。。

#include <iostream>#include <cstdio>#include <cstring>#include <stdlib.h>#include <algorithm>using namespace std;int a[1011][1011];int main(){    int n,m,k;    while(~scanf("%d%d%d",&n,&m,&k))    {        memset(a,0,sizeof(a));        int f=0;        int ss=0;        for(int i=0; i<k; i++)        {            int x,y;            scanf("%d%d",&x,&y);            if(f==0)            {                a[x][y]=1;                if(x-1>=1&&x-1<=n&&y-1>=1&&y-1<=m)                {                    if(a[x-1][y-1]&&a[x][y-1]&&a[x-1][y])                    {                        ss=i+1;                        f=1;                    }                }                if(x-1>=1&&x-1<=n&&y+1>=1&&y+1<=m)                {                    if(a[x-1][y] && a[x-1][y+1]&&a[x][y+1])                    {                        ss=i+1;                        f=1;                    }                }                if(x+1>=1&&x+1<=n&&y-1>=1&&y-1<=m)                {                    if(a[x][y-1] && a[x+1][y-1]&&a[x+1][y])                    {                        ss=i+1;                        f=1;                    }                }                if(x+1>=1&&x+1<=n&&y+1>=1&&y+1<=m)                {                    if(a[x][y+1] && a[x+1][y+1]&&a[x+1][y])                    {                        ss=i+1;                        f=1;                    }                }            }        }        printf("%d\n",ss);    }    return 0;}