C-C Radar Installation 解题报告

C-C    Radar Installation   解题报告

题目链接：http://acm.hust.edu.cn/vjudge/contest/view.action?cid=86640#problem/C

题目：

Description

 

 

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<tex2html_verbatim_mark> distance, so an island in the sea can be covered by a radius installation, if the distance between them is at most d<tex2html_verbatim_mark>     .

We use Cartesian coordinate system, defining the coasting is the x<tex2html_verbatim_mark>    -axis. The sea side is above x<tex2html_verbatim_mark>     -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<tex2html_verbatim_mark>      -       y<tex2html_verbatim_mark> coordinates.

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Input

The input consists of several test cases. The first line of each case contains two integers n<tex2html_verbatim_mark>(1n1000)<tex2html_verbatim_mark> and d<tex2html_verbatim_mark>      , where n<tex2html_verbatim_mark> is the number of islands in the sea and d<tex2html_verbatim_mark> is the distance of coverage of the radar installation. This is followed by n<tex2html_verbatim_mark> 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.

Output

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.

Sample Input

3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

Case 1: 2
Case 2: 1

题目大意：
在一条海岸线的一侧有若干个岛屿，将海岸线看成是X轴，X轴以上是大海，在海岸上安装雷达使其能覆盖全部岛屿，给出雷达的覆盖半径和岛屿位置，求最少使用多少
雷达才能将所有岛屿全部覆盖。无法覆盖输出-1.

分析：
1.可以换一种想法，求雷达覆盖小岛就是以小岛为圆心，雷达的覆盖半径为半径画圆
2.小岛到海岸线的距离大于半径，不能覆盖所有小岛，输出-1
3.若小于半径，该圆与X轴的左交点为line[i].l=(double)x-sqrt((double)d*d-y*y);右交点为 line[i].r=(double)x+sqrt((double)d*d-y*y);
4.将左交点排序，判断雷达位置，如果i点的左交点在当前雷达的右边，需要安装一个新雷达；如果i点的右交点在当前雷达的左边，则把当前雷达的圆心更新为该点的右交点

代码：

 #include<cstdio>
 #include<iostream>
 #include<cmath>
 #include<algorithm>
 using namespace std;

 const int maxn=;

 struct Line  //每个岛作半径为d的圆，与x轴所截的线段
 {
     double l,r;
 }line[maxn];

 bool cmp(Line a,Line b)//按照线段的左端点从小到大排序
 {
     return a.l<b.l;
 }

 int main()
 {
     int n,d;
     int i;
     int x,y;
     bool yes;//确定是不是有解
     int m=;
     while(scanf("%d%d",&n,&d)!=EOF)
     {
         yes=true;
         int cnt=;
         if(n==&&d==)  //n,d均为0，程序结束
             break;
         for(i=;i<n;i++)
         {
             scanf("%d%d",&x,&y);
             if(yes==false)
                 continue;
             if(y>d)
                 yes=false;
             else
             {
                 line[i].l=(double)x-sqrt((double)d*d-y*y);
                 line[i].r=(double)x+sqrt((double)d*d-y*y);
             }
         }
         if(yes==false)  //不能完全覆盖
         {
             printf("Case %d: -1\n",m++);
             continue;
         }
         sort(line,line+n,cmp);//排序
         cnt++;
         double now=line[].r;
         for(i=;i<n;i++)
         {
             if(line[i].r<now)  //与现在比较，这个很重要
                 now=line[i].r;
             else if(now<line[i].l)
             {
                 now=line[i].r;
                 cnt++;
             }
         }
         printf("Case %d: %d\n",m++,cnt);
     }

     return ;
 }