CodeForces 508A Pasha and Pixels

Pasha在他的手机上玩一款游戏,目标是在不形成2x2黑色像素方块的情况下涂色。玩家需要按照计划进行多步操作,每一步都会涂色一个像素。本篇介绍如何判断Pasha是否会输以及在哪一步形成禁止的2x2方块。

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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 m pixels 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.

Examples
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

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

int map[1111][1111];

int main(){	
	int n,m,k;
	int x,y;
	while(cin>>n>>m>>k){
		int index=0;
		memset(map,0,sizeof(map));
		for(int i=0;i<k;i++){
			scanf("%d %d",&x,&y);
			map[x][y]=1;
			if(map[x+1][y+1]==1&&map[x+1][y]==1&&map[x][y+1]==1&&index==0){
				index=i+1;
				continue;
			}
			if(map[x-1][y-1]==1&&map[x-1][y]==1&&map[x][y-1]==1&&index==0){
				index=i+1;
				continue;
			}
			if(map[x-1][y+1]==1&&map[x-1][y]==1&&map[x][y+1]==1&&index==0){
				index=i+1;
				continue;
			}
			if(map[x+1][y-1]==1&&map[x+1][y]==1&&map[x][y-1]==1&&index==0){
				index=i+1;
				continue;
			}
		}
		printf("%d\n",index);
	}
	return 0;
}

 
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