POJ 1328 Radar Installation

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 77074		Accepted: 17265

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

Beijing 2002

 

 

 

解析：每一个岛屿要么不能被雷达覆盖到，要么在海岸线上存在一个区间，该岛屿只能被放置在此区间内的雷达覆盖到。可以处理出这n个区间，区间重叠的部分可以共用一个雷达，接下来就很容易了。

 

 

#include <cstdio>
#include <algorithm>
#include <cmath>
using namespace std;

const int MAXN = 1000+5;
int n, d;
int x[MAXN], y[MAXN];
pair<double, double> p[MAXN];

void solve()
{
    double offset;
    for(int i = 0; i < n; ++i){
        if(d < y[i]){
            printf("-1
");
            return;
        }
        offset = sqrt(d*d-y[i]*y[i]);
        p[i].first = x[i]-offset;
        p[i].second = x[i]+offset;
    }
    sort(p, p+n);
    int res = 1;
    double l = p[0].first, r = p[0].second;
    for(int i = 1; i < n; ++i){
        if(p[i].first > r){
            ++res;
            l = p[i].first;
            r = p[i].second;
        }
        else{
            l = max(l, p[i].first);
            r = min(r, p[i].second);
        }
    }
    printf("%d
", res);
}

int main()
{
    int cn = 0;
    while(scanf("%d%d", &n, &d), n){
        for(int i = 0; i < n; ++i)
            scanf("%d%d", &x[i], &y[i]);
        printf("Case %d: ", ++cn);
        solve();
    }
    return 0;
}