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POJ 1328 Radar Installation

2024-08-11 22:19:54   218人阅读

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. 
技术分享 
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

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\n");            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\n", 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;}

　　

POJ 1328 Radar Installation


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