贪心-覆盖问题-圆心位置的判别

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. 

 
Figure A Sample Input of Radar Installations

 

Input

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 

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

题意：之前做的覆盖问题都是直接知道圆心的位置，这个是需要自己算的，题意就是给出一个x-y轴，x轴上方给出几个点，要求用圆把这些点包围起来，圆心必须在x轴上，并且圆的半径是固定的。给出多组测试用例，第一行两个数n,d;表示n个点，圆的半径是d,下面是n个点，两个数x,y,表示一个点的坐标位置。

题解：将结构体按照横坐标从小到大排序，算出圆心在横坐标上的位置，如果下一个圆心在这个圆心的左边，那么圆心就更新成为下一个圆心，如果下一个圆心在这个圆心的右边，就需要计算这个点到当前圆心的距离是否大于d，如果大于的话，就需要重新画个圆，画个图就比较容易理解

#include<stdio.h>
#include<string.h>
#include<math.h>
#include<algorithm>
using namespace std;
struct Ti
{
    int x,y;
}t[1010];
bool cmp(Ti a,Ti b)
{
    if(a.x!=b.x)
    return a.x<b.x;
    return a.y>b.y;
}
int main()
{
    int n,d,h=0;
    while(~scanf("%d%d",&n,&d))
    {
        h++;
        if(n==0&&d==0)
            break;
        int i,maxx=-1;
        for(i=0;i<n;i++)
        {
            scanf("%d %d",&t[i].x,&t[i].y);
            maxx=max(maxx,t[i].y);
        }
        if(maxx>d)
        {
            printf("Case %d: ",h);
            printf("-1\n");
            continue;
        }
        sort(t,t+n,cmp);
        double o=t[0].x+sqrt((double)d*d-t[0].y*t[0].y),o1;
        int sum=1;
        i=1;
        while(i<n)
        {
            o1=t[i].x+sqrt((double)d*d-t[i].y*t[i].y);
            if(o1<o)
            {
                o=o1;
                i++;
                continue;
            }
            if(sqrt((double)(t[i].x-o)*(t[i].x-o)+t[i].y*t[i].y)>d)
            {
                o=o1;
                sum++;
            }
            i++;
        }
        printf("Case %d: %d\n",h,sum);
    }
    return 0;
}