POJ1328 Radar Installation

POJ1328 Radar Installation

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

1
2
3
4
5
6
7
8
9
3 2
1 2
-3 1
2 1

1 2
0 2

0 0

Sample Output

1
2
Case 1: 2
Case 2: 1

Answer

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#include <cstdio>
#include <algorithm>
#include <cmath>
#include <utility>
using namespace std;

pair<double, double> islands[1010];

int main() {
int n, d;
int tc = 0;
while (true) {
bool flag = true;
scanf("%d %d", &n, &d);
if (n == 0 && d == 0)
break;
for (int i = 0; i < n; ++i) {
int x, y;
scanf("%d %d", &x, &y);
if (y > d) {
flag = false;
continue;
}
double t = d * d - y * y;
t = sqrt(t);
islands[i].first = -t + x;
islands[i].second = t + x;
}

printf("Case %d: ", ++tc);

if (!flag) {
printf("-1\n");
continue;
}

sort(islands, islands+n);
int cnt = 1;
double right = islands[0].second;
for (int i = 1; i < n; ++i) {
if (islands[i].second < right) {
right = islands[i].second;
} else if (islands[i].first > right) {
right = islands[i].second;
++cnt;
}
}


printf("%d\n", cnt);
}
return 0;
}

Analysis

  1. 给出 n 个岛和半径 d,岛的 y 坐标不小于 d,则必然失败。
  2. 求出每个岛和 x 的两(一)个交点,则有 n 个区间,转化为区间贪心选择问题。
  3. 将每个区间按照左端点的顺序增序排列。若下一个区间的右端点 < 此时的右端点,更新。否则若下一个区间的左端点大于此时的右端点,更新,答案+1。
  4. 几何问题:每一个岛为圆心,d 为半径画的圆与 x 轴两个交点的区间,其中任意一点为圆心,d 为半径都可以到岛。因为任意一点,都在岛为圆心的园内,所以岛圆心的距离必然小于 d。
# POJ

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