标签：
原题：
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.
Input
The input consists of several test cases. The first line of each case contains two integers n (1
n1000) 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 nlines 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
题意:
给出一个坐标轴, y的正半轴是海, 负半轴是大陆, x轴是海岸线;
然后在海上有很多海岛, 需要雷达监控, 现在的问题是, 至少需要多少个雷达, 能全部覆盖这些海岛?
分析：可以化解成区域选最少点问题，注意装换ok
代码：
#include<iostream> #include<iomanip> #include<cstdio> #include<cmath> #include<algorithm> #include<cstring> using namespace std; struct node { double x, y; bool operator <(node &b)const { return y<b.y || (y == b.y&&x>b.x); } }point[1000+10]; int main() { int n, d; double x, y; int kase = 0; while (cin >> n >> d&&n != 0 && d != 0) { kase++; int ok = 1; int cnt = 0; for (int i = 0; i < n; i++) { scanf("%lf%lf", &x, &y); if (y>d){ ok = 0; } else{ point[i].x = x - sqrt(d*d - y*y); point[i].y = x + sqrt(d*d - y*y); } } getchar(); sort(point, point + n); double end; for (int i = 0; i < n; i++) { if (i == 0){ end = point[i].y; continue; } if (end < point[i].x){ end = point[i].y; cnt++; } } if (ok == 0){ printf("Case %d: -1\n", kase); } else{ printf("Case %d: %d\n",kase, cnt + 1); } } return 0; }
UVALive - 2519 Radar Installation 解题心得
标签：
原文地址：http://www.cnblogs.com/shawn-ji/p/4711862.html
