POJ1328 Radar Installation 【贪心·区间选点】

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 54593		Accepted: 12292

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

Source

field=source&key=Beijing+2002" style="text-decoration:none">Beijing 2002

#include <stdio.h>
#include <string.h>
#include <math.h>
#include <algorithm>
 
#define maxn 1010
using namespace std;
 
struct Node {
    double u, v;
    friend bool operator<(const Node& a, const Node& b) {
        return a.u < b.u;
    }
} E[maxn];
int N, D;
 
int main() {
    int i, ok, id, ans, cas = 1;
    double x, y, d, flag;
    while(scanf("%d%d", &N, &D), N) {
        printf("Case %d: ", cas++);
        ok = 1; id = 0;
        for(i = 0; i < N; ++i) {
            scanf("%lf%lf", &x, &y);
            if(y > D) ok = 0;
            if(!ok) continue;
            d = sqrt(D * D - y * y);
            E[id].u = x - d;
            E[id++].v = x + d;
        }
 
        if(!ok) {
            printf("-1\n");
            continue;
        }
 
        sort(E, E + id);
 
        flag = E[0].v; ans = 1;
        for(i = 1; i < N; ++i) {
            if(E[i].u <= flag) {
                if(E[i].v <= flag) flag = E[i].v;
                continue;
            }
            ++ans; flag = E[i].v;
        }
 
        printf("%d\n", ans);
    }
    return 0;
}