【贪心算法】POJ-1328 区间问题
2017-08-04 14:51
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一、题目
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 1
2 1
1 2
0 2
0 0
Sample Output
Case 1: 2Case 2: 1
二、思路&心得
贪心算法与区间问题:以每个岛屿的坐标为中心,d为半径构造圆,该圆与X轴的两个交点即构成一个区间,若在这个区间上的任何雷达均可扫描到该岛屿。通过对n个岛屿进行处理,可得到n个区间,则问题转化成区间问题。每个区间a[i]有两个端点:first和second,对区间数组a按second进行升序排序,然后从左向右扫描,对于每一个区间a[i],若a[i].first小于之前选择的second的值,则不做任何处理,知道找到大于second的区间,然后进行下一个循环。
在数据输入的时候可进行特判,如若有岛屿的Y坐标大于d或则Y坐标<0或则d<0,则输入完成后直接返回-1即可。
数据定义记得用浮点型进行定义。
PS:在做贪心问题时,务必确定所做的贪心选择的正确性,在做这题时因为一开始的方向就是错误的,导致浪费了很多时间。
三、代码
#include<cstdio> #include<cmath> #include<algorithm> #define MAX_SIZE 1005 using namespace std; typedef pair<double, double> P; P a[MAX_SIZE]; int n, ans; double x, y, d; bool cmp(P a, P b) { if (a.second < b.second) return true; else return false; } int solve() { bool flag = true; double end; ans = 0; for (int i = 0; i < n; i ++) { scanf("%lf %lf", &x, &y); if (y > d || y < 0) flag = false; if (flag) { a[i].first = x - sqrt(d * d - y * y); a[i].second = x + sqrt(d * d - y * y); } } if (!flag || d < 0) return -1; sort(a, a + n, cmp); for (int i = 0; i < n; i ++) { if (flag) { end = a[i].second; ans ++; flag = false; continue; } if (a[i].first > end) { flag = true; i --; } } return ans; } int main() { int step = 1; while (~scanf("%d %lf", &n, &d)) { if (!n && !d) break; printf("Case %d: %d\n", step ++, solve()); getchar(); } return 0; }
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