POJ 1328 Radar Installation（区间贪心）

题意：x周上方是海面， 海面上有若干个小岛， 现在要在x轴上建立一些雷达站， 雷达站的覆盖面是一个圆，半径为d， 求最少用几个雷达覆盖所有小岛。

解题思路：

1.贪心法， 首先要把平面问题转化为一条线上的问题。即把一个雷达能这个岛的区间求出来，这样获得n个区间， 排序。

2.让最左边的岛放在第一个区间的最左边，用区间最右边的点去和下一个区间最左边的点去比较， 若是小于则建一个新的雷达站，并获得新雷达站区间最右边的点。若是最右边的点大于下一个区间最左边的点小于下一个区间最右边的点，则不用建立新雷达站，跟新最右边的点。如果最右边的点大于下一个区间最右边的点，则说明下一个区间包含于当前区间内，不用建立新的雷达站， 但是区间的最右端要等于下一个区间的最右端，即最右端要左移，因为在当前区间下可能还有被包含区间。举个例子三个区间可能是这样的

A：1 100

B：2    3

C：50 60

那么需要建立两个雷达站， 而不是一个。存在建立在B最右端的雷达站可以覆盖到A岛和B岛， 但覆盖不到C岛的情况。

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 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

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define MAX_N 1005
const int INF = 0x3f3f3f3f;

struct node
{
    double l;
    double r;
};

bool cmp(node a, node b)
{
    return a.l < b.l;
}

node t[MAX_N];

int main()
{
    int cas = 1;
    int n, d;
    int x, y;
    int ans = 0;
    while(scanf("%d%d", &n, &d) && (n + d))
    {
        ans = 0;
        for(int i = 0 ; i < n ; i++)
        {
            scanf("%d%d", &x, &y);
            if(y > d)
            {
                ans = -1;
                continue;
            }
            double tmp = sqrt(double(1.0 * d*d - 1.0 * y*y));
            t[i].l = 1.0 * x - tmp;
            t[i].r = 1.0 * x + tmp;
        }
        if(ans == -1)
        {
            printf("Case %d: %d\n", cas++, ans);
            continue;
        }
        sort(t, t + n, cmp);
        double r = t[0].r;
        ans = 1;
        for(int i = 1 ; i < n ; i++)
        {
            if( r < t[i].l)
            {
                ans++;
                r = t[i].r;
            }
            else if( r > t[i].r)
            {
                r = t[i].r;
            }
        }
        printf("Case %d: %d\n", cas++, ans);
    }

    return 0;
}