POJ 1328 Radar Installation

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

思路：典型贪心。算出每个岛在x轴上的左右区间，按照左区间排序。如某区间中完全包含另一个区间，则忽略这个区间。每个雷达都放在右区间处。
注意代码中的
return (((struct point*)a)->l)>(((struct point*)b)->l)?1:-1;
不能写成
return ((struct point*)a)->l)-(((struct point*)b)->l;

代码：

#include <cstdio>#include <iostream>#include <cstring>#include <vector>#include <map>#include <string>#include <cstring>#include <cmath>#include <algorithm>#include <cstdlib>using namespace std;struct point{    double r,l;}s[1000+10];int comp(const void *a,const void *b){    return (((struct point*)a)->l)>(((struct point*)b)->l)?1:-1;}int main(void){    #ifdef LOCAL        freopen("in.txt","rb",stdin);        //freopen("out.txt","wb",stdout);    #endif // LOCAL    int n,d,time=0;    while(scanf("%d%d",&n,&d)==2 && !(n==0 && d==0))    {        int ok=1;        for(int i=0;i<n;i++)        {            int tempx,tempy;            double tempd;            cin >> tempx >> tempy;            if(tempy>d) ok=0;            if(ok)            {                tempd=sqrt((double)(d*d-tempy*tempy));                s[i].l=tempx-tempd;                s[i].r=tempx+tempd;            }        }        if(!ok)        {            printf("Case %d: %d\n",++time,-1);            continue;        }        qsort(s,n,sizeof(s[0]),comp);        int vis[1000+10];        memset(vis,0,sizeof(vis));        for(int i=0;i<n;i++)            for(int j=i+1;j<n;j++)            {                if(s[j].l>s[i].r) break;                if(s[j].l>=s[i].l && s[j].r<=s[i].r)                {                    vis[i]=1;                    break;                }            }        int sum=0;        for(int i=0;i<n;i++)        {            if(!vis[i])            {                vis[i]=1;                sum++;                for(int j=i+1;j<n;j++)                {                    if(s[j].l<=s[i].r && s[i].r<=s[j].r)                        vis[j]=1;                    if(s[j].l>s[i].r) break;                }            }        }        printf("Case %d: %d\n",++time,sum);    }    return 0;}