AcWing 3293. 风险人群筛查
原题链接
简单
作者:
wjhhsa
,
2023-05-23 17:25:14
,
所有人可见
,
阅读 31
#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
int flag[25],g[25];
int main()
{
int n,k,t,x1,y1,x2,y2;
cin >> n>>k>>t>>x1>>y1>>x2>>y2;
for (int i = 1; i <= n; i ++ ){
int cnt = 0;
for(int j=1;j<=t;j++){
int x,y;
cin >>x>>y;
if((x>=x1&&x<=x2)&&(y>=y1&&y<=y2)) {
flag[i]=1;
cnt++;
if(cnt >= k) {
g[i]=1;
}
}
else {
cnt = 0;
}
}
}
int sum=0,gm=0;
for(int i=1;i<=n;i++){
sum+=flag[i],gm+=g[i];
}
cout << sum<<endl<<gm;
return 0;
}