题目
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
假设滑行是一条无限长的直线。陆地在海岸的一边，海洋在另一边。每个小岛都是位于海边的一个点。而任何位于海岸线上的雷达装置，只能覆盖d个距离，因此，如果它们之间的距离不超过d，海上的一个岛屿就可以被半径装置覆盖。
我们使用笛卡尔座标系，定义滑行是x轴。海侧高于x轴，陆侧低于x轴。给定海中每个岛屿的位置，以及雷达装置覆盖范围的距离，您的任务是编写一个程序，找出覆盖所有岛屿的雷达装置的最小数量。请注意，岛的位置由其x-y座标表示。
图A雷达装置的样本输入
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
输入由几个测试用例组成。每种情况的第一行包含两个整数n（1<=n<=1000）和d，其中n是海里的岛屿数，d是雷达装置的覆盖距离。接下来是n行，每行包含两个整数，表示每个岛位置的座标。接下来是一个空行来分隔这些案例。
输入端由包含一对零的行终止
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.
对于每个测试用例输出，一行由测试用例编号和所需的最少雷达安装数组成。”-1“安装意味着没有解决方案。
Sample Input
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2
Case 2: 1
题解
| 题解 |
|---|
| 把这个问题转换成区间问题 |
| 寻找有多少个没有交集的区间 |
| 如果有交集 则在第一个区间的最右添加雷达 |
| 根据我上面的图可以知道，最右添加是可以的 |
| 如果是有区间在第一个区间的里面 |
| 则在这个区间的最右 |
// #include <iostream>
#include <algorithm>
#include <vector>
#include <stdio.h>
#include <cmath>
using namespace std;
struct pair1
{
double first, second;
} p[1010];
bool cmp(pair1 e1, pair1 e2)
{
return e1.first < e2.first;
}
int main()
{
int N, R;
int cnt = 1;
while (~scanf("%d %d", &N, &R))
{
if (N == 0 && R == 0)
break;
int ans = 1;
for (int i = 0; i < N; i++)
{
double x, y;
scanf("%lf %lf", &x, &y);
// d是x 距离⚪在x轴上的交点的距离
double d = sqrt(R * R - y * y);
// 求出与x轴的两个交点
// 现在就转换成了区间问题
p[i].first = x - d;
p[i].second = x + d;
// 如果范围不够
if (y > R || y < 0 || R <= 0)
ans = -1;
}
// 转换成区间问题
sort(p, p + N, cmp);
// 第一个是要按装的
double t = p[0].second;
// 如果两个区间有交集 那么就可以覆盖的住
for (int i = 1; i < N && ans != -1; i++)
{
// 两个区间没有交集
if (p[i].first > t)
{
ans++;
t = p[i].second;
}
if (p[i].second < t)
{
t = p[i].second;
}
}
printf("Case %d: %d\n", cnt++, ans);
}
return 0;
}