• • 题目描述
  • 暴力遍历
    • 分析
    • 代码
    • 时间复杂度
  • 参考文章

题目传送门

题目描述

暴力遍历

分析

代码

#include<iostream>
#include<cstdio>
#include<vector>
using namespace std;
struct POI
{
	int x, y;
	int cnt;
	int score;
};
vector<POI> h;
int n;
int res[6];

bool check1(int x1, int y1, int x2, int y2)
{
	if(x2 == x1-1 && y2 == y1)   return true;
	if(x2 == x1+1 && y2 == y1)   return true;
	if(x2 == x1 && y2 == y1 - 1) return true;
	if(x2 == x1 && y2 == y1 + 1) return true;
	
	return false;
}

bool check2(int x1, int y1, int x2, int y2)
{
	if(x2 == x1 - 1 && y2 == y1 - 1) return true;
	if(x2 == x1 + 1 && y2 == y1 + 1) return true;
	if(x2 == x1 - 1 && y2 == y1 + 1) return true;
	if(x2 == x1 + 1 && y2 == y1 - 1) return true;
	
	return false;
}

int main()
{
	scanf("%d", &n);
	for(int i = 0; i < n; i++)
	{
		POI p;
		scanf("%d%d", &p.x, &p.y);
		p.cnt = 0;
		p.score = 0;
		h.push_back(p);
	}
	
	for(int i = 0; i < n; i++)
	{
		int x = h[i].x;
		int y = h[i].y;
		for(int j = 0; j < n; j++)
		{
			if(j == i) continue;
			
			int bx = h[j].x, by = h[j].y;
			
			if(check1(x, y, bx, by)) h[i].cnt++;
			if(check2(x, y, bx, by)) h[i].score++;
		}
	}
	for(int i = 0; i < n; i++)
	{
		if(h[i].cnt == 4)
			res[h[i].score]++;
	}
	for(int k = 0; k < 5; k++) printf("%d\n", res[k]);
	
	return 0; 
}

时间复杂度

参考文章

标签：x1,return,int,x2,201912,y1,y2,csp,回收站
来源： https://www.cnblogs.com/VanHa0101/p/16023424.html