題目
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
假設滑行是一條無限長的直線。陸地在海岸的一邊，海洋在另一邊。每個小島都是位於海邊的一個點。而任何位於海岸線上的雷達裝置，只能覆蓋d個距離，因此，如果它們之間的距離不超過d，海上的一個島嶼就可以被半徑裝置覆蓋。
我們使用笛卡爾座標系，定義滑行是x軸。海側高於x軸，陸側低於x軸。給定海中每個島嶼的位置，以及雷達裝置覆蓋範圍的距離，您的任務是編寫一個程序，找出覆蓋所有島嶼的雷達裝置的最小數量。請注意，島的位置由其x-y座標表示。
圖A雷達裝置的樣本輸入
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
輸入由幾個測試用例組成。每種情況的第一行包含兩個整數n（1<=n<=1000）和d，其中n是海里的島嶼數，d是雷達裝置的覆蓋距離。接下來是n行，每行包含兩個整數，表示每個島位置的座標。接下來是一個空行來分隔這些案例。
輸入端由包含一對零的行終止
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.
對於每個測試用例輸出，一行由測試用例編號和所需的最少雷達安裝數組成。”-1“安裝意味着沒有解決方案。
Sample Input
3 2
1 2
-3 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2
Case 2: 1
題解
| 題解 |
|---|
| 把這個問題轉換成區間問題 |
| 尋找有多少個沒有交集的區間 |
| 如果有交集 則在第一個區間的最右添加雷達 |
| 根據我上面的圖可以知道，最右添加是可以的 |
| 如果是有區間在第一個區間的裏面 |
| 則在這個區間的最右 |
// #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 距離⚪在x軸上的交點的距離
double d = sqrt(R * R - y * y);
// 求出與x軸的兩個交點
// 現在就轉換成了區間問題
p[i].first = x - d;
p[i].second = x + d;
// 如果範圍不夠
if (y > R || y < 0 || R <= 0)
ans = -1;
}
// 轉換成區間問題
sort(p, p + N, cmp);
// 第一個是要按裝的
double t = p[0].second;
// 如果兩個區間有交集 那麼就可以覆蓋的住
for (int i = 1; i < N && ans != -1; i++)
{
// 兩個區間沒有交集
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;
}