poj1328贪心

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 76566		Accepted: 17139

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

///////////////////////////////////////////////
分析：
１．以d为斜边作直角三角形，可知每个坐标在x轴上形成一个闭区间，在此区间内的雷达满足要求；
２．因此题目转化为求区间的集合数，相交的为一个集合，求最小集合数；
３．贪心规则：由左向右摆放雷达，因此从考虑最靠左的区间右端点开始，向右推进求集合数。

<span style="font-size:14px;">#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>
#include <cmath>
using namespace std;

int n;
int d;
pair<double , double>group[1000];
pair<signed int , signed int>coordinate[1000];
int times = 0;

int main(int argc, char const* argv[])
{
//freopen("input.txt","r",stdin);
while (scanf("%d%d",&n,&d)){
if((n == 0) && (d == 0)) break;
times++;
int ans = 1;
bool out = false;
if(d < 0){
out = true;
}
for (int i = 0; i < n; i += 1){
scanf("%d%d",&coordinate[i].first,&coordinate[i].second);
if (d >= coordinate[i].second){
double dd = sqrt(d*d-(coordinate[i].second)*(coordinate[i].second));
group[i].first = coordinate[i].first+dd;//group.right
group[i].second = coordinate[i].first-dd;//group.left
}
else{
out = true;
}
}
if(out){
printf("Case %d: -1\n",times);
continue;
}
sort(group,group+n);//depended on right
int i= 0;
for (int j = i+1; j < n; j += 1){
if (group[j].second > group[i].first){
ans++;
i = j;
}
}
printf("Case %d: %d\n",times,ans);
}
return 0;
}</span>