ZOJ-1360（POJ-1328） Radar Installation

Radar Installation
Time Limit: 2 Seconds      Memory Limit: 65536 KB
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.



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

Source: Asia 2002, Beijing (Mainland China)

————————————————————饥饿的分割线————————————————————
思路：一开始我的思路是自左至右尽量将雷达向右放，但是这么贪心是不对的。例如这组数据：
 4 5

-5 3

-3 5

 2 3

 3 3

答案是2。因为雷达放在[-3, 5]的正下方比放在[-5, 3]允许（安放雷达）的最右点更优。
那么转变思路。【--_--其实我是去看题解了……】
一个雷达能够覆盖一个岛屿当且仅当岛屿处于雷达范围内。但是雷达的位置不确定。不要紧，岛屿是确定的。岛屿能够被雷达覆盖，当且仅当“雷达处于岛屿范围内”。也就是说，岛屿允许的最左点和岛屿允许的最右点之间只要有一个雷达，岛屿就被覆盖了。那么要想让雷达尽量少，就要将雷达尽量放在重叠区间内。
所以将所有岛屿的坐标转化成区间。重叠的区间个数就是最少的雷达个数。这其实就是选择不相交区间问题。
友情提示：注意使用double。
代码如下：

/****************************************/
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <stack>
#include <queue>
#include <vector>
#include <map>
#include <string>
#include <iostream>
using namespace std;
/****************************************/
struct Node
{
double left, right;
}a[1010];

bool cmp(Node a, Node b)
{
if(a.right != b.right)
return a.right < b.right;
return a.left < b. left;
}

int main()
{
int n, d, cas = 1;
//freopen("rader.in", "r", stdin);
while(~scanf("%d%d", &n, &d), n||d) {
int maxi = -1, x, y;
for(int i = 0; i < n; i++) {
scanf("%d%d", &x, &y);
a[i].left = x - sqrt((double)d*d - y*y);
a[i].right = x + sqrt((double)d*d - y*y);
maxi = max(maxi, y);
}
printf("Case %d: ", cas++);
if(maxi > d) {
printf("-1\n");
}
else {
sort(a, a+n, cmp);
int ans = 1;
for(int i = 0, j = 1; j < n; j++) {
if(a[j].left > a[i].right) {
ans++;
i = j;
}
}
printf("%d\n", ans);
}
}
return 0;
}