poj1328

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 76641		Accepted: 17158

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

题目大意：求最小的雷达个数能够覆盖所有的岛屿,n为岛屿的个数,d时每个雷达能够覆盖的范围

思路：贪心，我们先对所有的岛屿按照x坐标排序,从左向右依次计算雷达在x轴的区间能够覆盖该岛屿。

如果发现区间有重叠，则更新区间范围，因为要同时满足覆盖两个岛屿，因此更新的区间就是重叠的区间

即区间左端点为两个大的一个，右端点为两个小的一个,如果不重叠则需要增加新的雷达.

代码如下：

#include <iostream>
#include<cstdio>
#include<cmath>
#include<algorithm>
#include<cstring>
using namespace std;
const int maxs = 1000+5;
const int INF = 0x3f3f3f3f;
struct Node
{
    int x,y;
}island[maxs];//岛屿
struct Area
{
    double x1,x2;//能够覆盖岛屿的雷达的左右两端点
}radar[maxs];
int n,d;

bool cmp(Node a,Node b)
{
    return a.x<b.x;
}
int main()
{
    //freopen("in.txt","r",stdin);
    int t = 1;
    while(scanf("%d%d",&n,&d)!=EOF&&n)
    {
        int ans = 0;
        memset(radar,0,sizeof(radar));
        radar[0].x1=radar[0].x2=-INF;
        for(int i=1;i<=n;i++)
            scanf("%d%d",&island[i].x,&island[i].y);
        sort(island+1,island+1+n,cmp);

        for(int i=1;i<=n;i++)
        {
            if(d<0||d<island[i].y||island[i].y<0)
            {
                ans = -1;
                break;
            }
            double x = sqrt(d*d-island[i].y*island[i].y);
            double x1 = island[i].x-x,x2=island[i].x+x;
            if(x1>radar[ans].x2)//说明需要加一个雷达
            {
                ans++;
                radar[ans].x1 = x1;
                radar[ans].x2 = x2;
            }
            else
            {
                //更正共用雷达的区间
                radar[ans].x1 = max(radar[ans].x1,x1);
                radar[ans].x2 = min(radar[ans].x2,x2);
            }
        }
        getchar();
        printf("Case %d: %d
",t++,ans);
    }
}