poj1328Radar Installation 贪心

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 64472 Accepted: 14497

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 21 2-3 12 11 20 20 0

Sample Output

Case 1: 2Case 2: 1

Source

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node{    double L,R;} p[1005];int cmp(node p1,node p2){    return p1.L<p2.L;}int main(){    int n,d,num=0;    while(cin>>n>>d)    {        num++;        if(n==0&&d==0)            break;        int flag=1;        for(int i=0; i<n; i++)        {            int u,v;            cin>>u>>v;            if(flag==0)    continue;            if(d<v)      //注意半径可以取负的,所以不能用d*d<v*v比较            {                flag=0;            }            else            {                p[i].L=(double)u-sqrt((double)(d*d-v*v));                p[i].R=(double)u+sqrt((double)(d*d-v*v));            }        }        if(flag==0)        {            printf("Case %d: -1\n",num);            continue;        }        sort(p,p+n,cmp);        double x=p[0].R;        int sum=1;        for(int i=1; i<n; i++)        {if(p[i].R<x)            {                x=p[i].R;            }            else if(x<p[i].L)            {                sum++;                x=p[i].R;            }        }        printf("Case %d: %d\n",num,sum);    }}/*把每个岛屿来当做雷达的圆心，半径为d，做圆，与x轴会产生两个焦点L和R，这就是一个区间；首先就是要把所有的区间找出来，然后x轴从左往右按L排序，再然后就是所谓的贪心把那些互相重叠的区间去掉就行了区间也就是雷达；*//* 3 -3 1 2 -3 2 2 1Case  ... -1; *///按R进行从左到右排序#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<algorithm>using namespace std;struct node{    double L,R;} p[1001];int cmp(node p1,node p2){    return p1.R<p2.R;}int main(){    int n,d,num=0;    while(cin>>n>>d)    {        num++;        if(n==0&&d==0)            break;        int flag=0;        for(int i=0; i<n; i++)        {            int u,v;            cin>>u>>v;            if(d<v)            {                flag=1;            }            else if(flag==0)            {                p[i].L=u-sqrt(d*d-v*v);                p[i].R=sqrt(d*d-v*v)+u;            }        }        if(flag)        {            printf("Case %d: -1\n",num);            continue;        }        sort(p,p+n,cmp);        double  xR=p[0].R;        double  xL=p[0].L;        int sum=1;        for(int i=1; i<n; i++)        {            if(p[i].L<=xR)            {            }            else if(p[i].L>xR)            {                xR=p[i].R;                sum++;            }        }        printf("Case %d: %d\n",num,sum);    }}