Radar Installation
| Time Limit: 1000MS | Memory Limit: 10000K |
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 0Sample Output
Case 1: 2 Case 2: 1
题意：给你n个岛和雷达的最大探测半径，雷达只能安装在x轴上，问最少用多少个雷达可以探测到所有的岛
解题思路: 每个雷达尽量多的覆盖，即在保证覆盖其已覆盖的岛的情况下，尽量向右多覆盖岛。只需要算出雷达要覆盖每个岛在x轴上的坐标区间，从左向右枚举所有区间，区间重复的岛屿用一个雷达即可。 如图 ，最少需要三个雷达分别放置在红色位置
ACCode
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>
#define inf 0x3f3f3f3f
using namespace std;
struct Node {
int x,y;
friend bool operator < (const Node t1,const Node t2) {
if(t1.x == t2.x)
return t1.y < t2.y;
return t1.x < t2.x;
}
};
Node loc[1005];
int main()
{
int n,d,Case=1;
while(scanf("%d%d",&n,&d) && (n || d)) {
bool flag = false;
for(int i=1;i<=n;i++) {
scanf("%d %d",&loc[i].x,&loc[i].y);
if(abs(loc[i].y) > d) flag = true;
}
if(flag) {
printf("Case %d: -1\n",Case++);
continue;
}
sort(loc+1,loc+1+n);
double r=inf,tl,tr;
int tot=1;
for(int i=1;i<=n;i++) {
tl = loc[i].x - double(sqrt(d*d-loc[i].y*loc[i].y)); // 当前岛屿要被探测到雷达可放置到最左端的位置
tr = loc[i].x + double(sqrt(d*d-loc[i].y*loc[i].y)); // 最右
if(tl > r) {
tot++;
r = tr;
}
else {
r = min(r,tr);
}
}
printf("Case %d: %d\n",Case++,tot);
}
return 0;
}