poj 1328 Radar Installation (贪心)

Radar Installation

Time Limit: 1000MS		Memory Limit: 10000K
Total Submissions: 48890		Accepted: 10917

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

给你n个点和雷达探测半径d，问在x轴上最少要放多少个雷达使雷达探测覆盖所有的点。

贪心思想：按x轴小到大排序，然后求得圆心，在判断过程中修正圆心位置，最后以O(n)求解。

//204K    16MS    C++    1223B    2014-04-27 09:22:25
#include<stdio.h>
#include<stdlib.h>
#include<math.h>
struct node{
    double x,y;     
}p[1005];
int cmp(const void*a,const void*b)
{
    node *x=(node*)a;
    node *y=(node*)b;
    if(x->x!=y->x) return (x->x)>(y->x)?1:-1;
    return (x->y)>(y->y)?-1:1;
}
double dis(node a,node b)
{
    return sqrt((a.x-b.x)*(a.x-b.x)+(a.y-b.y)*(a.y-b.y));
}
double getx(double d,node p0)
{
    return p0.x+sqrt(d*d-p0.y*p0.y);
}
int main(void)
{
    int n,k=1;
    int d0;
    while(scanf("%d%d",&n,&d0),n+d0)
    {
        double d=d0*1.0;
        int flag=0;
        for(int i=0;i<n;i++){
            scanf("%lf%lf",&p[i].x,&p[i].y);
            if(p[i].y>d) flag=1;
        }
        printf("Case %d: ",k++);
        if(flag){
            puts("-1");continue;
        }
        qsort(p,n,sizeof(p[0]),cmp);
        node q={getx(d,p[0]),0};
        int cnt=1;
        for(int i=1;i<n;i++){
            double temp=getx(d,p[i]);
            if(p[i].x<q.x && q.x>temp){ //开始少了这步.WA了几次 
                q.x=temp;continue;
            }
            if(dis(q,p[i])>d){
                cnt++;
                q.x=getx(d,p[i]);
            }
        }
        printf("%d\n",cnt);
    }
    return 0;
}