POJ 1328 Radar Installation（贪心）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 75586		Accepted: 16939

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

题目大意：
有一排海岸线，海岸线平行于X轴，海岸线的附近有很多岛屿，对于海岸线给定岛屿的坐标。问题是需要安装一些雷达来监测这些岛屿，每个雷达的监测半径为m，然后需要在海岸线上安装雷达，问你最少需要安装多少个，可以把所有的岛屿都监测到，输出数量，如果做不到，输出-1
题目解析：
根据给定的岛屿坐标，我们可以计算出每个岛屿在海岸线上安装雷达的区间，意思就是在这个区间内的任何地方安装雷达都是可以检测到此岛屿。这样我们根据n个岛屿的坐标就得到了n个不同的区间在海岸线上，然后我们只贪心的去处理这些区间，让每个区间内至少都存在一个雷达就可以了，那么从第一个区间， 选择最右边一个点， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后， 以前得到满足的区间， 现在仍然得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉，继续更新最右边的点，直到结束。
#include <algorithm>
#include <iostream>
#include <numeric>
#include <cstring>
#include <iomanip>
#include <string>
#include <vector>
#include <cstdio>
#include <queue>
#include <stack>
#include <cmath>
#include <map>
#include <set>
#define LL long long
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
const LL MAX = 405;
const double esp = 1e-6;
const double PI = 3.1415926535898;
const int INF = 0x3f3f3f3f;
using namespace std;

struct node{
double l,r;
}arr[1005];

bool cmp(node a,node b){
return a.l < b.l;
}

int main(){
int n,Cas = 1;
double m,k,nowx,nowy;
while(~scanf("%d %lf",&n,&m) && (n||m)){
int f = 0;
for(int i=0;i<n;i++){
scanf("%lf %lf",&nowx,&nowy);
if(nowy > m)
f = 1;
arr[i].l = nowx - sqrt(m * m - nowy * nowy);
arr[i].r = nowx + sqrt(m * m - nowy * nowy);
}
sort(arr,arr+n,cmp);
int ans = 1;
k = arr[0].r;
for(int i=0;i<n-1;i++){
if(arr[i+1].l > k){
k = arr[i+1].r;
ans += 1;
}
else if(arr[i+1].r < k){
k = arr[i+1].r;
}
}
if(f)
printf("Case %d: -1\n",Cas++);
else
printf("Case %d: %d\n",Cas++,ans);
}
return 0;
}