POJ 1328 Radar Installation (贪心)

Radar Installation


Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 82241 Accepted: 18418



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



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



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








AC 代码

#include <iostream>
using namespace std;

struct point
    double x,y;
    double s,e;
    bool operator<(const point &o)const
        return e<o.e;
} land[1005];

int solve(int n,int d)
    for(int i=0; i<n; i++)  //计算所有的线段
        double t=sqrt(d-land[i].y*land[i].y);
    sort(land,land+n);      //排序
    int ans=1;
    double r=land[0].e;
    for(int i=1; i<n; i++)
    return ans;

int main()
    int n,d,cat=1;
        bool flag=false;
        for(int i=0; i<n; i++)
            if(land[i].y>d)flag=true;   //是否可以检测到该岛屿
        if(flag)printf("Case %d: -1\n", cat++);
        else printf("Case %d: %d\n", cat++, solve(n,d*d));
    return 0;