一、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 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
二、思路&心得
- 贪心算法与区间问题：以每个岛屿的坐标为中心，d为半径构造圆，该圆与X轴的两个交点即构成一个区间，若在这个区间上的任何雷达均可扫描到该岛屿。通过对n个岛屿进行处理，可得到n个区间，则问题转化成区间问题。
- 每个区间a[i]有两个端点：first和second，对区间数组a按second进行升序排序，然后从左向右扫描，对于每一个区间a[i]，若a[i].first小于之前选择的second的值，则不做任何处理，知道找到大于second的区间，然后进行下一个循环。
- 在数据输入的时候可进行特判，如若有岛屿的Y坐标大于d或则Y坐标<0或则d<0，则输入完成后直接返回-1即可。
- 数据定义记得用浮点型进行定义。
- PS:在做贪心问题时，务必确定所做的贪心选择的正确性，在做这题时因为一开始的方向就是错误的，导致浪费了很多时间。
三、代码
#include<cstdio>
#include<cmath>
#include<algorithm>
#define MAX_SIZE 1005
using namespace std;
typedef pair<double, double> P;
P a[MAX_SIZE];
int n, ans;
double x, y, d;
bool cmp(P a, P b) {
if (a.second < b.second) return true;
else return false;
}
int solve() {
bool flag = true;
double end;
ans = 0;
for (int i = 0; i < n; i ++) {
scanf("%lf %lf", &x, &y);
if (y > d || y < 0) flag = false;
if (flag) {
a[i].first = x - sqrt(d * d - y * y);
a[i].second = x + sqrt(d * d - y * y);
}
}
if (!flag || d < 0) return -1;
sort(a, a + n, cmp);
for (int i = 0; i < n; i ++) {
if (flag) {
end = a[i].second;
ans ++;
flag = false;
continue;
}
if (a[i].first > end) {
flag = true;
i --;
}
}
return ans;
}
int main() {
int step = 1;
while (~scanf("%d %lf", &n, &d)) {
if (!n && !d) break;
printf("Case %d: %d
", step ++, solve());
getchar();
}
return 0;
}