Radar Installation
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Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 77760 Accepted: 17398
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.
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 21 2-3 12 11 20 20 0
Sample Output
Case 1: 2Case 2: 1
Source
Beijing 2002
#include<iostream>#include<stdio.h>#include<stdlib.h>#include<math.h>#include<algorithm>using namespace std;struct node{ double l; double r;} a[1010];bool comp(node a,node b){ if(a.r<b.r) return true; else if(a.r==b.r) { if(a.l>b.l) return true; return false; } return false;}int main(){ int num; int i,n; int cas=1; int x,d,y; double temp; while(~scanf("%d %d",&n,&d)&&(n||d)) { int flag=1; num=1; for(i=0; i<n; i++) { scanf("%d %d",&x,&y); temp=(double)(d*d-y*y); if(temp<0||d<0) { flag=0; } else { a[i].l=x-sqrt((double)d*d-y*y); a[i].r=x+sqrt((double)d*d-y*y); } } if(!flag) { printf("Case %d: -1\n",cas++); continue; } else { sort(a,a+n,comp); temp=a[0].r; for(i=1; i<n; i++) { if(a[i].l>temp) { temp=a[i].r; num++; } } printf("Case %d: %d\n",cas++,num); } } return 0;}
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- Radar Installation
- Radar Installation
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- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
- Radar Installation
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