• POJ 1328


    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

    #include <cmath>
    #include <iostream>
    #include <cstdio>
    #include <cstring>
    #include <cstdlib>
    #include <vector>
    #include <algorithm>
    using namespace std;
    
    struct Space{
        double l, r;
        Space(double l = 0, double r = 0):l(l),r(r){}
        bool operator < (const Space & rhs) const {
            return r < rhs.r;
        }
    };
    Space a[1024];
    
    int main()
    {
        ios::sync_with_stdio(false);
        int n, kase = 0, ok = 1;
        double d;
        while(memset(a, 0, sizeof(a)), ok = 1, cin >> n >> d && (n || d)){
            for(int i = 0; i < n; ++i){
                double x, y; cin >> x >> y;
                if(y > d){ok = 0; continue;}
                double r = sqrt(d*d - y*y);
                a[i] = Space(x - r, x + r);
            }
            if(!ok){printf("Case %d: %d
    ", ++kase, -1); continue;}
            sort(a, a + n);
            double End = a[0].r;
            int cnt = 1;
            for(int i = 1; i < n; ++i){
                if(a[i].l > End ){
                    ++cnt;
                    End = a[i].r;
                }
            }
            printf("Case %d: %d
    ", ++kase, cnt);
        }
        return 0;
    }


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  • 原文地址:https://www.cnblogs.com/kunsoft/p/5312688.html
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