POJ 1328：Radar Installation

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 50843 Accepted: 11415

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

思路：该题题意是为了求出能够覆盖所有岛屿的最小雷达数目，每个小岛对应x轴上的一个区间，在这个区间内的任何一个点放置雷达，则可以覆盖该小岛，区间范围的计算用[x-sqrt(d*d-y*y),x+sqrt(d*d-y*y)];这样，问题即转化为已知一定数量的区间，求最小数量的点，使得每个区间内斗至少存在一个点。每次迭代对于第一个区间， 选择最右边一个点， 因为它可以让较多区间得到满足， 如果不选择第一个区间最右一个点（选择前面的点）， 那么把它换成最右的点之后， 以前得到满足的区间， 现在仍然得到满足， 所以第一个区间的最右一个点为贪婪选择， 选择该点之后， 将得到满足的区间删掉， 进行下一步迭代， 直到结束。

以左区间为准：

#include<cstdio>#include<cstring>#include<algorithm>#include<iostream>#include<cmath>using namespace std;const int M = 1000 + 5;double x, y;double dis;           //圆的半径double temp;int ans;             //用于统计结果int n;               //点的个数int flag;            //用于判断struct node{    double left;       //左区间    double right;     //右区间 } island[M];bool cmp(node a, node b)   //按左区间从小到大排序{    return a.left<b.left;}int main(){    int cas=0;    while(scanf("%d%lf", &n, &dis) && n && dis)    {        flag=0;        ans=0;        cas++;        for(int i=0; i<n; i++)        {            scanf("%lf%lf", &x, &y);            if(y>dis||y<0)               //不符合的点                flag=1;            island[i].right = x+sqrt(dis*dis-y*y);                  island[i].left = x-sqrt(dis*dis-y*y);        }        if(flag==1)            printf("Case %d: -1\n",cas);        else        {            sort(island, island+n, cmp);   //sort排序            temp = island[0].right;                   ans=1;            for(int i=1; i<n; i++)            {                if(island[i].right<temp)          //说明这个区间被上个区间所包含，只要小区间满足的点，大区间肯定也满足                    temp=island[i].right;                else if(island[i].left>temp)     //说明这两个区间没有重合部分，也就不能用一个圆表示，则结果加一                {                    ans++;                    temp=island[i].right;                }            }            printf("Case %d: %d\n", cas, ans);        }    }    return 0;}

一右区间为准：（这是我同学写的右区间）

#include<iostream>#include<cstring>#include<cmath>#include<cstdio>#include<algorithm>using namespace std;struct node{double a,b;}s[1005],l[1005];int cmp(node x,node y){return x.b<y.b;}int main(){int n,m,k=0;while(cin>>n>>m){if(n==0&&m==0)   break;int i,j,p=0;for(i=0;i<n;i++){scanf("%lf%lf",&s[i].a,&s[i].b);if(s[i].b<0||s[i].b>m)p=1;l[i].a=s[i].a-sqrt(m*m-s[i].b*s[i].b);l[i].b=s[i].a+sqrt(m*m-s[i].b*s[i].b);}sort(l,l+n,cmp);int sum=1; double r=l[0].b;for(i=1;i<n;i++){if(l[i].a>r){r=l[i].b;sum++;}}printf("Case %d: ",++k);if(p)cout<<-1<<endl;elsecout<<sum<<endl;}return 0;}