POJ 1328 Radar Installation (贪心+线段覆盖)

Radar Installation

Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77865 Accepted: 17424

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

题意：海上有n个岛，x轴上的雷达的范围为d, 问覆盖所有岛屿最少需要多少雷达。

如果abs(y)>d,则返回-1.

思路：以岛为中心作半径为d的圆，交x轴于两点成一条线段，排序各个线段，若第i个线段的l>第i-1个线段的r  则说明前一个雷达并不能覆盖此岛屿，需要新建设雷达。

代码好像大概会RE。但是思想正确。。就这么放着吧- -

#include <cstdio>  #include <cstring>  #include <cmath>  #include <cstdlib>  #include <algorithm>  #include <queue>  #include <stack>  #include <map>  #include <vector>#include <iostream>#define INF 0x3f3f3f3f  #define eps 1e-4  #define MAXN (1000+10)  #define MAXM (1000000)  #define CLR(a, b) memset(a, (b), sizeof(a))  #define MOD 100000007  #define LL long long  #define lson node<<1, l, mid  #define rson node<<1|1, mid+1, r   using namespace std;int xy[MAXN][MAXN];struct line{double l, r;}l[MAXN],temp;//sort函数需要重载“<”，以升序排列bool operator < (line a, line b) {return a.l < b.l;}int main() {int n, d,cnt=1;double x, y;while (scanf("%d%d",&n,&d) && (n || d)) {bool flag = true;for (int i = 1; i <= n; ++i) {scanf("%d%d", &x, &y);if (abs(y) > d) flag = false;else {l[i].l = x*1.0 - sqrt(d*d - y*y);l[i].r = x*1.0 + sqrt(d*d - y*y);}}cout << "Case " << cnt++ << ": ";if (!flag) {cout << -1 << endl;}else {temp = l[1];int ans = 1;sort(l, l + n);for (int i = 2; i <= n; ++i) {if (l[i].l > temp.r) {ans++;temp = l[i];}else if (temp.r > l[i].r) {temp = l[i];}}cout << ans << endl;}}return 0;}