poj_1328_Radar Installation(贪心)

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

思路

题意：用最少的雷达覆盖所有的岛屿。
本来想从雷达入手去做，可是好多东西是变的，没法解决，也比较麻烦。上网找了题解之后是从岛屿入手。因为雷达都安在一条线上，只有x的变化，而且雷达的覆盖范围一定，这就转化成了类似区间覆盖问题。
雷达覆盖半径以确定，通过岛屿（x,y）位置可以找出最大区间，（以雷达覆盖半径为半径，岛屿为圆心画圆，即可得到区间）将区间的左端点从小到大排序，判断岛屿是否在范围内即可。
如果下一个区间的左端点与上一个区间的右端点不重合，则雷达数+1，并更新右端点，如果下一个区间完全被上一个包含在内，则更新右端点。

用结构体：区间左右端点
sort排序函数：按左端点从小到大，若左端点相等，则按右端点从小到大

代码

#include <iostream>#include<algorithm>#include<math.h>using namespace std;struct zuobiao{    double x;  //区间左    double y;  //区间右};bool cmp(zuobiao a , zuobiao b ){    if(a.x!=b.x)        return a.x<b.x;    return a.y<b.y;}int main(){    struct zuobiao s[1005];    int n,d,count,num=1;    int i,x,y;    double t,p;    while(cin>>n>>d)    {        count=1;        if(n==0&&d==0)            break;        for(i=0;i<n;i++)        {            cin>>x>>y;            if(y>d)                count=-1;            t=sqrt(d*d-y*y);//解直角三角形            s[i].x=x-t;//区间左端点            s[i].y=x+t;//区间右端点        }        if(count!=-1)        {            sort(s,s+n,cmp);            p=s[0].y;//区间右端点            for(i=1;i<n;i++)            {                if(s[i].x>p)//下一个区间左端点与上一个区间右端点没有重合                {                    count++;                    p=s[i].y;//更新右端点                }                else if(s[i].y<p)//下一个区间完全被上一个包含在内                    p=s[i].y;            }        }        cout<<"Case "<<num<<": "<<count<<endl;        num++;    }    return 0;}