【贪心】Radar Installation（POJ1328）

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

Sample Output

Case 1: 2
Case 2: 1

分析

刚开始我是对雷达分析，不能想出全部覆盖小岛的方法，所以改对小岛分析，小岛是必须被覆盖的，所以可以得到覆盖小岛的雷达的横坐标最小值和最大值，然后按横坐标最大值从小到大排序，运用贪心算法，在横坐标最小值处放雷达，雷达就可以尽可能多的覆盖小岛。还有两个特殊情况：

①小岛y坐标大于雷达覆盖半径的话，就要输出-1。

②当横坐标最大值相同时，应让半径小的排在前，因为要在横坐标最小值处放雷达，这样可以保证横坐标最大值相同时都可以覆盖到。

参考代码

//这是我写的，大多数样例都过了，但是一直WA，T.T

#include <iostream>
#include <cstdio>
#include <math.h>
#include <algorithm>
using namespace std;

struct  P{
    double l;
    double r;
    }p[1001];

bool compare(const P& a,const P& b);
int main()
{
    int sum=1;
    int n;
    double d;
    double x,y;
    int maxy=0;
    int num=1;
    while(1){
        maxy=0;
        sum=1;
       scanf("%d%lf",&n,&d);
       if(n==0&&d==0)
        break;
       for(int i=0;i<n;i++){
        scanf("%lf%lf",&x,&y);
        if(y>maxy){
            maxy=y;
        }
        double t=sqrt(d*d-y*y);
        p[i].l=x-t;
        p[i].r=x+t;
       }

       if(maxy>d){
        printf("Case %d: -1
",num++);
       }

       sort(p,p+n,compare);
       int minx=p[0].r;
       for(int i=1;i<n;i++){
            if(minx<p[i].l){
                sum++;
                minx=p[i].r;
            }

       }
       if(maxy<=d)
       printf("Case %d: %d
",num++,sum);
    }



    return 0;
}
bool compare(const P& a,const P& b){
    if(a.r==b.r)
        return a.l>b.l;
    return a.r<b.r;
}

//AC代码摘自http://www.cnblogs.com/q-c-y/p/4713557.html

#include<iostream>
#include<cmath>
#include<cstdio>
#include<algorithm>
#define MAX 1005
using namespace std;
struct note{
    double left,right;
}b[MAX];
double cmp(note a,note b)      //结构体排序
{
    if(a.right==b.right) return a.left>b.left; 
        else
    return a.right<b.right;
}
int acount,n; 
void search(double maxn)      //找到最少的雷达数
{
    int i;
    for(i=1;i<n;i++)
        if(b[i].left>maxn)
        {
            maxn=b[i].right;        //都取最右端
            acount++;
        }
}
int main()
{
    double d,x,y;
    scanf("%d%lf",&n,&d);
    int t=0;
    while(n||d)
    {
        t++;
        int o=0;
        acount=0;
        for(int i=0;i<n;i++)
        {
            scanf("%lf%lf",&x,&y);
            if(y>d)                      //判断是否可以扫射到岛
                o++;
            b[i].left=x-sqrt((d*d)-(y*y));         //预处理，变成区间
            b[i].right=x+sqrt((d*d)-(y*y));
        }
        if(!o)
        {
            sort(b,b+n,cmp);
            search(b[0].right);
            printf("Case %d: %d
",t,acount+1);
        }
        else
            printf("Case %d: -1
",t);
        scanf("%d%lf",&n,&d);
    }
    return 0;
}