NYOJ 287 Radar 贪心之 区间选点

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

题意：给定点集S={（xi，yi）i=1.2.3...n}，求用圆心在x轴上，半径为r的圆覆盖S所需的圆的最少个数。

解题思路：先把给定的xi，yi,r转化为x轴上的区间，即圆心所在的区间，这样就转化为了区间选点问题。先对右端点从小到大排序，右端点相同时，左端点从小到大排序。

代码：

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

struct radar
{
    double a, b;
}r[1005];

bool comp(radar a1, radar a2)
{
    if(a1.b != a2.b) //对右端点从小到大排序
        return a1.b < a2.b;
    return a1.a < a2.a; //右端点相同，左端点从小到大排序
}

int main()
{
    int n, d, i, cas = 0;
    while(~scanf("%d%d",&n,&d) && (n + d))
    {
        double x, y;
        int flag = 0, m = 0;
        for(i = 0; i < n; i++)
        {
            scanf("%lf%lf",&x,&y);
            if(fabs(y) > d)
            {
                flag = 1; //有覆盖不到的点
                continue;
            }
            double diff = sqrt(d * d - y * y);
            r[m].a = x - diff;
            r[m++].b = x + diff;
        }
        printf("Case %d: ",++cas);
        if(flag)
        {
            printf("-1\n");
            continue;
        }
        sort(r, r + m, comp);
        int cnt = 1, p = 0;
        for(i = 1; i < m; i++)
        {
            if(r[i].a <= r[p].b)
                continue;
            else
            {
                cnt++;
                p = i;
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}