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poj 1328贪心

2024-08-12 00:10:19   220人阅读

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

 

 

 

大概题义就是要在x轴上放圆心，使得放最少的圆可以把所有点覆盖，输出圆的个数

#include <iostream>#include <string.h>#include <math.h>#include <stdio.h>#include <algorithm>using namespace std;#define Maxn 10010struct Node{    double a,b;}A[Maxn];bool cmp(Node a,Node b){    if(a.a != b.a){        return a.a < b.a;    }else{        return a.b < b.b;    }}double wenbao(double R,double high){    return sqrt( pow(R,2.0) - pow(high,2.0) );}int main(){    int N;    double R;    double a,b;    int count = 1;    while(cin >> N >> R,N+R){        bool flag = true;        for(int i = 0; i < N; i++){            // cin >> a >> b;            scanf("%lf%lf",&a,&b);//用scanf,否则会超时            if(fabs(b) > R){                flag = false;            }else{                double temp = wenbao(R,b);                A[i].a = a - temp;                A[i].b = a + temp;            }        }        sort(A,A+N,cmp);        Node p = A[0];        int cnt = 1;        for(int i = 1; i < N; i++){            if(A[i].a > p.b){                cnt++;                p = A[i];            }else if(A[i].b < p.b){//请仔细思考                p = A[i];            }        }        if(flag){            printf("Case %d: %d\n",count++,cnt);        }else{            printf("Case %d: -1\n",count++);        }    }}

　　

poj 1328贪心


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