POJ1328 Radar Installation (贪心)

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 87179 Accepted: 19538

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

Beijing 2002

思路：这道题目的难点在于如何把给定的已知条件，转化成相对熟悉的方法来求解。已知存在雷达和岛屿，要求用最少的雷达数目来覆盖全部的岛屿，如果存在不能够覆盖掉的，则输出-1，否则输出需要的最少的雷达数目。题目中岛屿的个数以及坐标都是固定的，要求的是雷达的数目，所以就可以以岛屿的坐标为圆心，以雷达扫描半径为圆的半径，画圆，然后求出圆和x轴的交点，如果两个圆和x轴的交点有重合，则说明这两个岛屿可以共用一个雷达，所以只需要求出以岛屿为圆心的圆与x轴的两个交点即可，以这两个交点为区间，然后求区间的重叠即可，

AC代码如下：

#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;int n,d;struct node{    int x,y;    double disl,disr;}p[1005];bool cmp(node A, node B){    if(A.disl == B.disl)        return A.disr < B.disr;    return A.disl < B.disl;}int main(){    int ans = 1;    while(~scanf("%d%d",&n,&d)){        if(n == 0 && d == 0)            break;        int flag = 0;        for(int i = 0; i < n; i ++){            scanf("%d%d",&p[i].x,&p[i].y);            if(p[i].y > d)//判断是否存在大于雷达半径的岛屿                flag = 1;        }        if(flag){//存在大于雷达半径的岛屿，直接输出-1即可            printf("Case %d: -1\n",ans++);            continue;        }        for(int i = 0; i < n; i ++){            double dist = sqrt(d * d - p[i].y * p[i].y);//勾股定理求以该岛屿为圆心与x轴交点的横坐标            p[i].disl = -dist + p[i].x;            p[i].disr = dist + p[i].x;//求当前区间的左右端点        }        sort(p,p+n,cmp);//按照左端点进行排序        int cnt = 1;        double r = p[0].disr;//记录第一个区间的右端点        for(int i = 1; i < n; i ++){            if(p[i].disl > r){//当前区间的左端点大于以前某区间的右端点，则说明不重叠，雷达数目+1，更新右端点                cnt ++;                r = p[i].disr;            }            else if(p[i].disr < r){//跟新右端点，使得右端点的值最小，因为只有在某一区间内相交的区间最多，雷达数目才能够最少                r = p[i].disr;            }        }        printf("Case %d: %d\n",ans++,cnt);    }    return 0;}