CodeForces - 203B

最新推荐文章于 2019-07-24 20:09:15 发布

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该问题描述了一个游戏场景，Valera在n×n的棋盘上依次涂黑m个格子，询问最少需要多少步可以形成一个边长为3的黑色正方形。输入包括棋盘大小n和涂黑格子数m，以及每步涂黑的坐标。输出是最小的形成3x3黑色方块的步数，若无法形成则输出-1。

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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.

Examples

Input

Output

Input

Output

-1

题意：

给一个n和m，代表有n*n的方格，m行输入给的格子，问到第几行时会出现3*3的方格都被走过；

先把每一步的路都存起来，再暴力找哪个是3*3的，并找出这个3*3中最大的是第几步；

#include <iostream>
#include <string.h>
#include<stdio.h>
#include<algorithm>
#include<math.h>
using namespace std;
int maze[1005][1005];
int ji[1005][1005];
bool re(int i,int j)//判断
{
    if(maze[i][j]==1)
        if(maze[i-1][j]==1)
            if(maze[i+1][j]==1)
                if(maze[i][j-1]==1)
                    if(maze[i][j+1]==1)
                        if(maze[i-1][j-1]==1)
                            if(maze[i-1][j+1]==1)
                                if(maze[i+1][j-1]==1)
                                    if(maze[i+1][j+1]==1)
                                        return true;
    return false;
}
int main()
{
    int n,m;
    while(~scanf("%d %d",&n,&m))
    {
        memset(maze,0,sizeof(maze));
        memset(ji,0,sizeof(ji));
        int flag=0;
        for(int i=0; i<m; i++)
        {
            int a,b;
            scanf("%d %d",&a,&b);
            ji[a][b]=i+1;
            maze[a][b]=1;
        }
        int ans=1000005,nnn;
        for(int i=1; i<=n; i++)
        {
            for(int j=1; j<=n; j++)
            {
                if(re(i,j))
                {
                    flag=1;
                    nnn=ji[i][j];
//                    if(ji[i][j]>nnn)nnn=ji[i][j];
                    if(ji[i-1][j]>nnn)nnn=ji[i-1][j];
                    if(ji[i+1][j]>nnn)nnn=ji[i+1][j];
                    if(ji[i][j-1]>nnn)nnn=ji[i][j-1];
                    if(ji[i][j+1]>nnn)nnn=ji[i][j+1];
                    if(ji[i-1][j-1]>nnn)nnn=ji[i-1][j-1];
                    if(ji[i-1][j+1]>nnn)nnn=ji[i-1][j+1];
                    if(ji[i+1][j-1]>nnn)nnn=ji[i+1][j-1];
                    if(ji[i+1][j+1]>nnn)nnn=ji[i+1][j+1];
                    if(nnn<ans)
                        ans=nnn;
                }
            }
        }
        if(flag)
            printf("%d\n",ans);
        else
            printf("-1\n");
    }
    return 0;

}