POJ 1328 Radar Installation(贪心——最少区间选点覆盖)

Radar Installation

Time Limit: 1000MS Memory Limit: 10000K

Total Submissions: 93549 Accepted: 20884

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

Source

Beijing 2002

题意 给你N个点的直角坐标，用半径为d，圆心在x轴的圆包括它们 求最小的圆的个数

做法：根据勾股定理求出每个点在x轴上能被覆盖的范围，求这么多区间最少的点覆盖

1、对这些区间按左端点排序

2、取每次覆盖区间的最右能覆盖到的端点tmp

对下一个区间有三种情况



1 2情况都可覆盖到，但是1的情况是右端点缩小了，就要把tmp赋值为b

3的情况是这个区间和当前的覆盖区间无交集，就要开一个新的区间了

PS : 一开始tmp声明成了Int死都过不了 gg

#include <cstdio>
#include <iostream>
#include <cstring>
#include <cmath>
#include <algorithm>
#include <queue>
#include <map>
#include <stack>
#include <vector>
#define maxn 10010
#define maxe 100010
typedef long long ll;
using namespace std;
const double eps=1e-5;
const int inf=0x3f3f3f3f3f;
struct Node
{
double x,y;
double a,b;
bool init(double d)
{
double t=d*d-y*y;
if(t<0)return false;
double c=sqrt(t);
a=x-c;
b=x+c;
return true;
}
}node[maxn];
bool cmp(Node s,Node r)
{
return s.a<r.a;
}
int main()
{
int n;
double d;
int cas=1;
//freopen("in.txt","r",stdin);
int flag=0;
while(scanf("%d%lf",&n,&d)!=EOF)
{
if(n==0&&d==0)break;
flag=0;

if(d<0)flag=1;
for(int i=0;i<n;i++)
{
scanf("%lf%lf",&node[i].x,&node[i].y);
if(!node[i].init(d))flag=1;
else if(abs(node[i].y)>d)flag=1;
}
printf("Case %d: ",cas++);
if(flag){puts("-1");continue;}
sort(node,node+n,cmp);
double tmp = node[0].b;
int ans=1;
for(int i=1;i<n;i++)
{
if(tmp>node[i].b)tmp=node[i].b;
else if(tmp<node[i].a)
{
ans++;
tmp=node[i].b;
}
}
printf("%d\n",ans);

}
return 0;
}