#POJ1328#Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 88545 Accepted: 19868

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

题意：

给定海岛个数、雷达半径以及各海岛坐标，求能覆盖所有海岛的最小雷达数雷达只能建在岸上。

这题的关键在确定范围上，即每个岛能在哪些范围被覆盖到，而自己的思维总是局限在多大的圆，怎么放才能覆盖更多的岛

确定出每个岛能被覆盖的范围，按右端点排序，从左向右遍历，去找左端点在覆盖内的最大的右端点，然后继续。

Code:

StatusAcceptedTime32msMemory208kBLength1140LangC++Submitted2017-07-13 13:33:06Shared

#include<iostream>#include<cstdio>#include<cstdlib>#include<cmath>#include<algorithm>using namespace std;const int Max = 10000;const double eps = 0.0001;struct node{double x, y;bool operator < (const node & X)const{return y < X.y;}}Joint[Max];int N, D;int X[Max], Y[Max];bool  getint(int & num){    char c;    int flg = 1;    num = 0;    while((c = getchar()) < '0' || c > '9')    if(c == '-')    flg = -1;    while(c >= '0' && c <= '9')    {    num = num * 10 + c - 48;    c = getchar();}    num *= flg;return 1;}double Get_Dis(int a, int b){return sqrt(1.0 * b * b - 1.0 * a * a);}int main(){int tot = 0;while(getint(N) && getint(D) && N && D){int Ans = 0;for(int i = 1; i <= N; ++ i){getint(X[i]), getint(Y[i]);double tmp = Get_Dis(Y[i], D);Joint[i].x = X[i] - tmp,Joint[i].y = X[i] + tmp;if(Y[i] > D || D <= 0 || Y[i] < 0)Ans = -1;}sort(Joint + 1, Joint + 1 + N);for(int i = 1; i <= N && Ans != -1; ){int j;for(j = i + 1; j <= N && Joint[j].x <= Joint[i].y + eps; ++ j);++ Ans;i = j;}printf("Case %d: %d\n", ++ tot, Ans);}return 0;}