题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
using namespace std;
const int N = 1010;
struct Node
{
    int x;
    int y;
    int flag;
    int count;
}node[N];

int cnt[5];
int n;

bool fl(int x1,int y1,int x2,int y2)
{
    if((x2 == x1 && y2 == y1 - 1) || (x2 == x1 && y2 == y1 + 1) || (x2 == x1 + 1 && y2 == y1) || (x2 == x1 - 1 && y2 == y1))
    {
        return true;
    }
    else
    {
        return false;
    }
}

bool f(int x1,int y1,int x2, int y2)
{
    if((x2 == x1 - 1 && y2 == y1 - 1) || (x2 == x1 - 1 && y2 == y1 + 1) || (x2 == x1 + 1 && y2 == y1 + 1) || (x2 == x1 + 1 && y2 == y1 - 1))
    {
        return true;
    }
    else
    {
        return false;
    }
}

int main()
{
    cin>>n;
    for(int i = 1;i <= n;i++)
    {
        int x,y;
        cin>>x>>y;
        node[i].x = x;
        node[i].y = y;
        node[i].count = 0;
    }
    for(int i = 1;i <= n;i++)
    {
        int re = 0;
        for(int j = 1;j <= n;j++)
        {
            if(fl(node[i].x, node[i].y, node[j].x, node[j].y))
            {
                if(re == 4)
                {
                    break;
                }
                re++;
            }
        }
        if(re == 4)
        {
            node[i].flag = 1;
        }
    }
    for(int i = 1;i <= n;i++)
    {
        for(int j = 1;j <= n;j++)
        {
            if(node[i].flag)
            {
                if(f(node[i].x, node[i].y, node[j].x, node[j].y))
                {
                    node[i].count++;
                }
            }
        }
    }
    for(int i = 1;i <= n;i++)
    {
        if(node[i].flag)
        {
            cnt[node[i].count]++;
        }
    }
    for(int i = 0;i < 5;i++)
    {
        cout<<cnt[i]<<endl;
    }
}