<cf>Game on Paper

最新推荐文章于 2020-07-25 13:31:53 发布

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最新推荐文章于 2020-07-25 13:31:53 发布

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文章标签： numbers input output

本文链接：https://blog.youkuaiyun.com/cs_zlg/article/details/7749031

本文介绍了一种算法挑战，任务是在一个n×n的网格中找出最小步数使至少出现一个3x3的全黑方格。提供了两种解决策略及其实现代码。

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 x_i, y_i (1 ≤ x_i, y_i ≤ 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 test(s)

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

input
4 12
1 1
1 2
1 3
2 2
2 3
1 4
2 4
3 4
3 2
4 2
4 1
3 1

output
-1如果一个九宫格都被涂黑，姑且称为“黑九宫格”，任意一个九宫格的九个格分别记为：中，上，下，左，右，左上，右上，左下，右下。第一种思路：对于输入的每个方格坐标（x，y），假设这个方格∈黑九宫格，若（x，y）为中格，考查以(x,y)为中格的九宫格是否全为黑，若(x,y)为下格，考查以(x+1,y)为中格的九宫格是否全黑，以此类推，至多执行九次就可判定输入的(x,y)是否∈黑九宫格。AC Code：#include <stdio.h>
#define N 10001
bool a[N][N]={0};
int m,n,x,y,i,count;

bool check(int x,int y)
{
    if(x==0||y==0)
        return false;
    else
        return a[x][y-1]&&a[x][y]&&a[x][y+1]&&a[x-1][y]&&a[x-1][y-1]&&a[x-1][y+1]&&a[x+1][y-1]&&a[x+1][y]&&a[x+1][y+1];

}

int main()
{
    scanf("%d%d",&n,&m);
    for(i=1;m--;i++)
    {
        scanf("%d%d",&x,&y);
        a[x][y]=true;
        if(check(x,y)||check(x,y+1)||check(x,y-1)||check(x+1,y)||check(x+1,y-1)||check(x+1,y+1)
           ||check(x-1,y)||check(x-1,y+1)||check(x-1,y-1))
           {
               printf("%d\n",i);
               return 0;
           }
    }
    printf("-1\n");
    return 0;
}

思路二：这个方法较之上面的巧得多！上面的是蛮力，这个是智取。每输入一个坐标(x,y)，就对以其为中格的九个格各做一次+1操作，直到出现一个格子对应的值为9（就是+1操作了九次），就能判定有黑九宫格，且这个格子∈黑九宫格（但不一定是中格）。在纸上画出一个九宫格就可看出若输入的（x1，y1）能使(x2,y2)进行+1操作,(x1,y1)只能∈以(x2,y2)为中格的九宫格，故而某格值为9，则以该格为中格的九宫格全黑。AC Code#include<stdio.h>
int s[1020][1020],n,m,x,y;
int main()
{
	scanf("%d %d",&n,&m);
	for(int i=0;i<m;i++)
	{
		scanf("%d %d",&x,&y);
		for(int j=x-1;j<=x+1;j++)
			for(int k=y-1;k<=y+1;k++)
			{
				s[j][k]++;
				if(s[j][k]==9)
				{
				    printf("%d",i+1);
				    return 0;
				}
			}
	}
	puts("-1");
	return 0;
}