POJ——1328 Radar Installation（区间覆盖问题）

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 65881		Accepted: 14767

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

区间覆盖问题，想了好久，根据所给的坐标判断在x轴上的覆盖范围，排序，然后用贪心思想解决。

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<math.h>
#include<ctype.h>
#include<stdlib.h>
#include<string>
#include<algorithm>
#include<vector>
#include<set>
#include<map>
#include<list>
#include<queue>
#include<stack>
#include<iomanip>
#include<numeric>
#include <istream>　　　　 //基本输入流
#include <ostream>　　　　 //基本输出流
#include <sstream>　　　　 //基于字符串的流
#include <utility>　　　　 //STL 通用模板类
#include <complex.h>　　 //复数处理
#include <fenv.h>　　　　//浮点环境
#include <inttypes.h>　　//整数格式转换
#include <stdbool.h>　　 //布尔环境
#include <stdint.h>　　　//整型环境
#include <tgmath.h>　　　//通用类型数学宏
#define L(a,b,c) for(int a = b;a >= c;a --)
#define M(a,b,c) for(int a = b;a < c;a ++)
#define N(a,b) memset(a,b,sizeof(a));
const int INF=1<<31;
const int inf=-INF;
typedef int T;
typedef double D;
typedef char C;

using namespace std;

struct node
{
    double x;
    double y;
} s[1010],temp;

bool comp (node x,node y)
{
    return x.x<y.x;
}

int main()
{
    int n,num=0;
    double a,b;
    double r;
    while(~scanf("%d%lf",&n,&r))
    {
        if(n==0&&r==0)
            break;
        bool flag=0;
        M(i,0,n)
        {
            cin>>a>>b;
            if(fabs(b)>r)           ///若大于r，则说明该点不能被覆盖
                flag=1;
            else
            {
                s[i].x=a*1.0-sqrt(r*r-b*b);
                s[i].y=a*1.0+sqrt(r*r-b*b);
            }
        }
        printf("Case %d: ",++num);
        if(flag)
        {
            printf("-1\n");
        }
        else
        {                       ///贪心思想
            int count=1;
            sort(s,s+n,comp);
            temp=s[0];
            M(i,1,n)
            {
                if(temp.y<s[i].x)
                {
                    ++ count;
                    temp=s[i];
                }
                else
                    temp.y=min(temp.y,s[i].y);      ///找出覆盖的最小区间
            }
            printf("%d\n",count);
        }
    }
    return 0;
}