POJ-1328（Radar Installation）

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.
InputThe 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.

Input

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

有多个测试样例每一个测试样例输入n和d表示后面有n个岛屿，和每一个雷达的覆盖半径
每一个岛屿用其坐标（x，y）表示，输入时没有括号和逗号，雷达只能放置在x坐标轴上，求用最少的雷达覆盖所有的岛屿，输出…（无法解决输出-1）
当然还要判断d是否大于0，不大于0当然无法解决…

AC代码（用vector）

#include<algorithm>
#include<iostream>
#include<math.h>
#include<vector>
#include<stdio.h>
using namespace std;
struct node
{
    double left;
    double right;
};
bool cmp(node a,node b)
{
    if(a.right!=b.right)
        return a.right<b.right;
    return a.left<b.left;
}
int main()
{
    int n,t=0;
    double d,x,y;
    vector<node> region;
    //vector<node>::iterator it;
    while(scanf("%d %lf",&n,&d))
    {
        if(n==0&&d==0)
            break;
        ++t;
        node now;
        bool judge=false;
        for(int i=0; i<n; ++i)
        {
            scanf("%lf %lf",&x,&y);
            if(y>d)
            {
                judge=true;
            }
            else
            {
                now.left=x-sqrt(d*d-y*y);
                now.right=x+sqrt(d*d-y*y);
                region.push_back(now);
            }
        }
        if(judge||d<0)
        {
            printf("Case %d: -1\n",t);
            continue;
        }
        sort(region.begin(),region.end(),cmp);
        int num=1;
        if(n>1)
            for(int i=1,j=0; i<n; ++i)
            {
                if(region[i].left>region[j].right)
                {
                    ++num;
                    j=i;
                }
            }
        printf("Case %d: %d\n",t,num);
        region.clear();
    }
    return 0;
}

AC代码（用数组）

#include<cstdio>
#include<math.h>
#include<algorithm>
using namespace std;
struct node
{
    double be;
    double en;
};
double cmp(node a, node b)
{
    if(a.en == b.en)
        return a.be > b.be;
        return a.en < b.en;
}
node wo[10000];
int main()
{
    int n;
    int code=0;
    double d;
    while(scanf("%d %lf",&n,&d))
    {
        int error=0;
        if(n==0&&d==0)
            break;
        code++;
        for(int i=0; i<n; i++)
        {
            double x,y;
            scanf("%lf %lf",&x,&y);
            if(y>d)
                error=1;
            else
            {
                wo[i].be=x-sqrt(d*d-y*y);
                wo[i].en=x+sqrt(d*d-y*y);
            }
        }
        if(error==1||d<0)
        {
            printf("Case %d: -1\n",code);
            continue;
        }
        sort(wo,wo+n,cmp);
        int num=1;
        if(n>1)
        {
            for(int i=1,j=0; i<n; i++)
            {
                if(wo[i].be>wo[j].en)
                {
                    j=i;
                    num++;
                }
            }
        }
        printf("Case %d: %d\n",code,num);
    }
    return 0;
}