POJ:1328 Radar Installation (贪心）

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 76845 Accepted: 17213

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

#include <cstdio>  #include <cstdlib>  #include <cstring>  #include <cmath>  #include <algorithm>  using namespace std;struct node//存放范围的左右横坐标（端点）{double left, right;} range[1100];int cmp(node a, node b)//以左端点的大小排序{return a.left < b.left;}int x[1100], y[1100];//存放岛的坐标int main(){int i, n, d, num, flag, s = 0;double z, p;while (scanf("%d%d", &n, &d) != EOF){if (n == 0 && d == 0) break;s++;flag = 0;for (i = 0; i<n; i++){scanf("%d%d", &x[i], &y[i]);if (y[i]>d)flag = 1;}if (flag)printf("Case %d: -1\n", s);else{for (i = 0; i<n; i++)//计算每个岛可以放雷达的范围{z = sqrt(d*d*1.0 - y[i] * y[i] * 1.0);range[i].left = (double)x[i] - z;range[i].right = (double)x[i] + z;}sort(range, range + n, cmp);//以每个范围的左端点大小进行排序num = 0;p = range[0].left-1;for (i = 0; i<n; i++)//贪心{if (range[i].left>p)//左端点大于上一个岛的可放雷达的范围的右端点，则需要新雷达{num++;p = range[i].right;}else if (range[i].right<p){p = range[i].right;//把雷达放到这个范围的右端点，以便与下一个范围的左端点比较大小}printf("Case %d: %d\n", s, num);}}return 0;}

转载地址：http://blog.csdn.net/deepmindman/article/details/52278572

转载地址：http://blog.csdn.net/deepmindman/article/details/52278572  刚做的时候想到的是思路一，这个思路是不对的。