CCF 201912-2 回收站选址

#include<iostream>
#include<map>
#include<string.h> 
using namespace std;

const int N = 1000;
int count[5];
pair<int, int> p[N];

int main()
{
	int n;
	map<pair<int, int>, int> ps;
	cin >> n;
	
	for(int i = 0; i < n; i ++)
	{
		int x, y;
		scanf("%d%d", &x, &y);
		
		p[i] = make_pair(x, y);
		ps[p[i]] = 1;
	}
	
	memset(count, 0, sizeof(count));
	
	for(int i = 0; i < n; i ++)
	{	
		int x = p[i].first;
		int y = p[i].second;
		
		if(ps[make_pair(x-1,y)] && ps[make_pair(x+1,y)] && ps[make_pair(x,y+1)] && ps[make_pair(x,y-1)])
			count[ps[make_pair(x-1,y+1)] + ps[make_pair(x+1,y+1)] + ps[make_pair(x-1,y-1)] + ps[make_pair(x+1,y-1)]] ++;
	}
	
	for(int i = 0; i < 5; i ++)
	{
		printf("%d\n", count[i]);
	}
	
	return 0;
}

参考了老师的博客，呜呜呜，老师的题解太棒了~
----------------------------------更新------------------------------------

#include <iostream>
#include <map>
using namespace std;

const int N = 1000 + 5;

map<pair<int, int>, bool> mp;
pair<int, int> p[N];
int a[5] = {0};

int main()
{
	int n, x, y, cnt;
	cin >> n;
	for(int i = 0; i < n; i ++)
	{
		cin >> x >> y;
		p[i].first = x;
		p[i].second = y;
		mp[p[i]] = true;
	}
	
	for(int i = 0; i < n; i ++)
	{ 
		cnt = 0;//每次判断都要清零 
		x = p[i].first;
		y = p[i].second;
		if(mp[make_pair(x,y+1)] && mp[make_pair(x,y-1)] && mp[make_pair(x-1,y)] && mp[make_pair(x+1,y)])
		{
			if(mp[make_pair(x+1,y+1)]) cnt ++;
			if(mp[make_pair(x+1,y-1)]) cnt ++;
			if(mp[make_pair(x-1,y+1)]) cnt ++;
			if(mp[make_pair(x-1,y-1)]) cnt ++;
			a[cnt] ++;
		}
	}
	
	for(int i = 0; i < 5; i ++)
	{
		cout << a[i] << endl;
	}
	return 0;
}