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 2
1 2
-3 1
2 1 1 2
0 2 0 0

Sample Output

Case 1: 2
Case 2: 1

分析:

简单的贪心,先按x轴排序,记录圆心位置范围,如果一个点的圆心范围和前面一个圆心范围有交集,就把前一个圆心范围更新为他们的交集

不然答案+1,将当前圆心范围记录下来,最后输出ans

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cmath>
using namespace std;
struct node {
int x, y;
};
node g[1111];
double posr[1111], posl[1111];
int n, d, ans;
int cmp (node a, node b) {
if (a.x == b.x) return a.y > b.y;
return a.x < b.x;
}
double fx (node x) {
double k = sqrt (1.*d * d - x.y * x.y);
return k;
}
int main() {
for (int t = 1; cin >> n >> d; t++) {
if (n == 0 && d == 0) break;
ans = 0;
for (int i = 1; i <= n; i++) {
cin >> g[i].x >> g[i].y;
if (g[i].y > d) ans = -1;
}
if (ans == 0) {
sort (g + 1, g + 1 + n, cmp);
if (n >= 1) {
posl[++ans] = g[1].x - fx (g[1]);
posr[ans] = g[1].x + fx (g[1]);
}
for (int i = 2; i <= n; i++) {
if (g[i].x == g[i - 1].x) continue;
if (g[i].x - fx (g[i]) > posr[ans]) {
posl[++ans] = g[i].x - fx (g[i]), posr[ans] = g[i].x + fx (g[i]);
continue;
}
posl[ans] = max (posl[ans], g[i].x - fx (g[i]) ), posr[ans] = min (posr[ans], g[i].x + fx (g[i]) );
}
}
printf ("Case %d: %d\n", t, ans);
}
return 0;
}
http://www.cnblogs.com/keam37/ keam所有 转载请注明出处

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