202009-2 风险人群筛查

http://118.190.20.162/view.page?gpid=T112
第一次提交 60分，显示超时

#include <iostream>
#include <cstdio>
#include <cstring>
#include <math.h>

using namespace std;

const int N = 22;
const int K = 1000;

struct point{
    int x;
    int y;
};
point points[N][K];
int times[N];
//bool visit[N];
int n, k, t, xl, yd, xr, yu;

int main(){
    scanf("%d %d %d %d %d %d %d", &n, &k, &t, &xl, &yd, &xr, &yu);
    point tmp;
    memset(times, 0, sizeof(times));
    //memset(visit, false, sizeof(visit));
    int pass = 0;
    int stay = 0;
    for(int i = 0; i < n; i++){
        int maxt = 0;
        for(int j = 0; j < 2 * t; j += 2){
            cin >> tmp.x >> tmp.y;
            points[i][j/2] = tmp;
            if(tmp.x >= xl && tmp.x <= xr && tmp.y >= yd && tmp.y <= yu){
                times[i]++;
                if(maxt < times[i]) maxt = times[i];
            }else{
                times[i] = 0;
            }
        }
        if(maxt >= k){
            stay++;
            pass++;
            //visit[i] == true;
            //printf("%d\n", i);
        }else if(maxt > 0){
            pass++;
            //printf("%d\n", i);
        }    
    }
    printf("%d\n%d\n", pass, stay);
    return 0;
}