HDOJ 4007 Dave【最大覆盖集】

一时失言乱红尘 2021-12-10 16:32 212阅读 0赞

# Dave #

**Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65768/65768 K (Java/Others)  
Total Submission(s): 2768    Accepted Submission(s): 926**

Problem Description

Recently, Dave is boring, so he often walks around. He finds that some places are too crowded, for example, the ground. He couldn't help to think of the disasters happening recently. Crowded place is not safe. He knows there are N (1<=N<=1000) people on the ground. Now he wants to know how many people will be in a square with the length of R (1<=R<=1000000000). (Including boundary).

Input

The input contains several cases. For each case there are two positive integers N and R, and then N lines follow. Each gives the (x, y) (1<=x, y<=1000000000) coordinates of people.

Output

Output the largest number of people in a square with the length of R.

Sample Input

3 2 1 1 2 2 3 3

Sample Output

*Hint*

If two people stand in one place, they are embracing.

Source

[ The 36th ACM/ICPC Asia Regional Dalian Site —— Online Contest ][The 36th ACM_ICPC Asia Regional Dalian Site _ Online Contest]

AC代码：

1 #include <stdio.h>
     2 #include <string.h>
     3 #include <algorithm>
     4 #include <math.h>
     5 using namespace std;
     6 struct node{
     7     int x,y;
     8 }a[1010],b[1010];
     9 bool cmp(node a,node b){
    10     return a.x < b.x;
    11 }
    12 int main(){
    13     int y[1010];
    14     int n,r,i,j,Max;
    15     int x1,x2,y1,y2;
    16     int x_min,x_max,y_min,y_max;
    17     while(EOF != scanf("%d%d",&n,&r)){
    18         Max = -1;
    19         x_min = y_min = 999999999;
    20         x_max = y_max = -1;
    21         for(i=0;i<n;i++){
    22             scanf("%d%d",&a[i].x,&a[i].y);
    23             if(a[i].x < x_min)
    24                 x_min = a[i].x;
    25             if(a[i].x > x_max)
    26                 x_max = a[i].x;
    27             if(a[i].y < y_min)
    28                 y_min = a[i].y;
    29             if(a[i].y > y_max)
    30                 y_max = a[i].y;
    31 
    32             b[i] = a[i];
    33         }
    34         if(y_max-y_min <= r && x_max-x_min <= r){
    35             printf("%d\n",n);
    36             continue;
    37         }
    38         sort(a,a+n,cmp);
    39         int Max = 0;
    40         for(int i=0;i<n;i++){
    41             int k = 0;
    42             for(int j = i;a[j].x <= a[i].x + r && j < n;j++)//对x值不大于a[j].x + r遍历
    43                 y[k++] = a[j].y;//将比a[i] 的 x值小的a[j]点 的y值加入y数组
    44             sort(y,y+k);//对y数组排序
    45             int flag = 0,temp = 0;
    46             for(int j = 0;j < k && temp < k ;j++){
         //对y数组中所有元素遍历
    47                 while(y[temp] - y[j] <= r && temp < k)
    48                     temp++;
    49                 if(temp -j > Max)
    50                     Max= temp - j;
    51             }
    52         }
    53         printf("%d\n",Max);
    54     }
    55     return 0;
    56 }

转载于:https://www.cnblogs.com/wushuaiyi/p/3638687.html

[The 36th ACM_ICPC Asia Regional Dalian Site _ Online Contest]: http://acm.hdu.edu.cn/search.php?field=problem&key=The%2036th%20ACM/ICPC%20Asia%20Regional%20Dalian%20Site%20%E2%80%94%E2%80%94%20Online%20Contest&source=1&searchmode=source