poj1328 greedy radar, land, island issues

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 60381		Accepted: 13610

Description

Assume the coasting is a infinite straight line. Side of coasting, sea in the other. Each of small island was a point locating in the sea side. and any radar installation, locating on the coasting, can only cover D-distance, so a island in the sea can is 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 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 a 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 all case contains-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 was followed by n lines each containing and 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 all test case output one line consisting of the test case number followed by the minimal number of radar installation S needed. "-1" installation means no solution for this case.

Sample Input

3 21 2-3 12 11 20 20 0

Sample Output

Case 1:2case 2:1

Source

Beijing 2002Test Instructions: There is an axis on the x axis above the sea is the land x-axis is the coastline, the Sea has n small islands, now to install radar on the coastline,

Radar coverage is a circle with a radius of R, please cover all the islets with a minimum of radar; output-1 when unable to overwrite

Idea: Calculate the center of the radar where each island can be covered, that is, a circle with the radius of the small island as the center of R, the intersection of the circle and the x-axis:

The left intersection is X-SQRT (r*r-y*y); The right intersection is X+SQRT (r*r-y*y);

Sort by the left node, if the left intersection of I points is on the right side of the current radar, a new radar will be installed to update the radar

Otherwise, if the right intersection point of I is also on the left side of the current radar, the center of the current radar is updated to the right intersection point;

Code

：：

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace Std;
struct node{
Double X;
Double y;
}QUE[1005];
BOOL CMP (struct node a,struct node B) {
Return a.x<b.x;
}
int main () {
int n;
Double D;
int count=0;
while (scanf ("%d%lf", &n,&d)!=eof) {
int sum;
if (n==0&&d==0)
Break
int flag=0;
count++;
for (int i=0;i<n;i++) {
Double A, B;
scanf ("%lf%lf", &a,&b);
QUE[I].X=A-SQRT (D*D-B*B);
QUE[I].Y=A+SQRT (D*D-B*B);
if (b>d| | d<=0| | B&LT;0)
flag=1;

}
if (!flag) {
Sort (que,que+n,cmp);
Double temp;
TEMP=QUE[0].Y;
Sum=1;
for (int i=1;i<n;i++) {
if (que[i].x>temp) {
sum++;
TEMP=QUE[I].Y;
}
else if (que[i].y<temp) {
TEMP=QUE[I].Y;
}
}
}
if (flag)
printf ("Case%d: -1\n", count);
Else
printf ("Case%d:%d\n", count,sum);
}
return 0;
}

poj1328 greedy radar, land, island issues