poj 1328 Radar Installation 【贪心】【区间选点问题】
Radar Installation
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 54798 | Accepted: 12352 |
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
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.
![](http://poj.org/images/1328_1.jpg)
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
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 2 1 2 -3 1 2 1 1 2 0 2 0 0
Sample Output
Case 1: 2 Case 2: 1
题意:就是找最少的站,来覆盖所有的点。
思路:我们能够以点来做半径为d的圆,与x轴的相交,假设不相交那么肯定完不成任务,反之就转化成了区间选点问题。
代码:
#include <stdio.h> #include <string.h> #include <algorithm> #include <math.h> using namespace std; #define M 1005 struct node { double st, en; }s[M]; int cmp(node a, node b){ if(a.en == b.en) return a.st > b.st; return a.en<b.en; } int main(){ int n, v = 1; double d; while(scanf("%d%lf", &n, &d), n||d){ int i, j; double a, b; int flag = 0; for(i = 0; i < n; i ++){ scanf("%lf%lf", &a, &b); if(b>d) flag = 1; if(flag == 0){ s[i].en = a+sqrt(d*d-b*b); s[i].st = a-sqrt(d*d-b*b); //printf("%lf %lf %d..\n", s[i].st, s[i].en, i); } // scanf("%lf%lf", &s[i].st, &s[i].en); } printf("Case %d: ", v++); if(flag){ printf("-1\n"); continue; } sort(s, s+n, cmp); int ans = 1; double maxr = s[0].en; i = 1, j = 0; while(i < n){ if(s[i].st <= s[j].en){ //if(maxr < s[i].en) maxr = s[i].en; ++i; } else { //if(j == i-1) j = i; ++ans; // maxr = s[i].en; } } printf("%d\n", ans); } return 0; }