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  • NYOJ 287 Radar

    Radar

    时间限制:1000 ms  |           内存限制:65535 KB
    难度:3
     
    描述
    Assume the coasting is an infinite straight line. Land is in one side of coasting, sea in the other. Each small island is a point locating in the sea side. And any radar installation, locating on the coasting, can only cover d distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d.

    We use Cartesian coordinate system, defining the coasting is the x-axis. The sea side is above x-axis, and the land side below. Given the position of each island in the sea, and given the distance of the coverage of the radar installation, your task is to write a program to find the minimal number of radar installations to cover all the islands. Note that the position of an island is represented by its x-y coordinates.

     

     
    输入
    The input consists of several test cases. The first line of each case contains two integers n (1<=n<=1000) and d, where n is the number of islands in the sea and d is the distance of coverage of the radar installation. This is followed by n lines each containing two integers representing the coordinate of the position of each island. Then a blank line follows to separate the cases.
    The input is terminated by a line containing pair of zeros
    输出
    For each test case output one line consisting of the test case number followed by the minimal number of radar installations needed. "-1" installation means no solution for that case.
    样例输入
    3 2
    1 2
    -3 1
    2 1
    
    1 2
    0 2
    
    0 0
    样例输出
    Case 1: 2
    Case 2: 1
    思路:确定岛在x轴上可覆盖区间(即在此范围内,信号可覆盖岛)右边界排序;
    代码:
     1  
     2 #include<stdio.h>
     3 #include<math.h>
     4 #include<algorithm>
     5 using namespace std;
     6 const int MAXN=1005;
     7 struct Line
     8 {
     9     double l,r;
    10 }line[MAXN];//每个岛作半径为d的圆,截得区间
    11 bool cmp(Line a,Line b)
    12 {
    13     return a.r<b.r;
    14 }
    15 int main()
    16 {
    17     int n,d;
    18     int i;
    19     int x,y;
    20     int num=1,flag,count;
    21     while(scanf("%d %d",&n,&d)&&n&&d)
    22     {
    23         flag=1;
    24         for(i=0;i<n;i++)
    25         {
    26             scanf("%d %d",&x,&y);
    27             if(!flag) continue;
    28             if(y<=d){
    29               line[i].l=(double)x-sqrt((double)d*d-y*y);
    30                line[i].r=(double)x+sqrt((double)d*d-y*y);
    31             }else flag=0;
    32         }
    33         if(!flag)
    34         {
    35             printf("Case %d: -1
    ",num++);
    36             continue;
    37         }
    38         sort(line,line+n,cmp);
    39         count=1;
    40         double now=line[0].r;
    41         for(i=1;i<n;i++)
    42         {
    43 
    44             if(line[i].l<=now+0.00005) continue;
    45             now=line[i].r;
    46             count++;
    47         }
    48       printf("Case %d: %d
    ",num++,count);
    49     }
    50     return 0;
    51 }
    52         
    View Code
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  • 原文地址:https://www.cnblogs.com/luoshuihanbing/p/3288599.html
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