POJ1328解题报告
来源:互联网 发布:华为软件开发云 编辑:程序博客网 时间:2024/05/16 17:28
Radar Installation
Time Limit: 1000MS Memory Limit: 10000KTotal Submissions: 40799 Accepted: 9034
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
题意:假设海岸线是一条无限延伸的直线。陆地在海岸线的一侧,而海洋在另一侧。每一个小的岛屿是海洋上的一个点。雷达坐落于海岸线上,只能覆盖d距离,所以如果小岛能够被覆盖到的话,它们之间的距离最多为d。
题目要求计算出能够覆盖给出的所有岛屿的最少雷达数目。
我们假设岛屿i它的x坐标为island[i][0],而y坐标为island[i][1],那么有以下几种情况是invalide的,即输出-1的情况:
1.island[i][1]<0
2.abs(island[i][1])<d
3.d<0
其他的情况,应该就是正常情况,进入计算最小雷达数目。
如上图,红色的点为岛屿,那么能够覆盖到此岛屿的雷达所在的区间,应该就是以该岛屿为圆心的圆与x轴交点所在的区间。
这样,我们就可以计算出所有岛屿的雷达所在的区间,得到一个区间数组。
我们将这个数组按照区间左部分进行排序,那么重叠部分就表明这些岛屿的雷达可以共用一个。从而计算出最终解。
代码如下:
#include<stdio.h>#include<limits.h>#include<math.h>#define MAXNUM 12701//以岛屿为圆心,以d为半径做圆;得到各个圆与x轴相交的区间;去掉重复区间,即得到雷达数目 int calMin(int **island,int n,int d);int main() { int num = 0; char s[10]; while(1){ num++; int n,d,i; scanf("%d%d",&n,&d); if(n==0&&d==0){ break; } int **island = (int**)malloc(sizeof(int*)*n); for(i=0;i<n;i++){ island[i] = (int*)malloc(sizeof(int)*2); scanf("%d%d",&island[i][0],&island[i][1]); } //计算最少雷达个数 int min = calMin(island,n,d); printf("Case %d: %d\n",num,min); //读取空白行 gets(s); free(island); } system("pause"); return 0;}int calMin(int **island,int n,int d){ double *arr = (double*)malloc(sizeof(double)*n*2); int i,j; for(i=0;i<n;i++){ if(island[i][1]<0||island[i][1]>d||d<0){ return -1; } //计算左右区间 if(abs(island[i][1]-d)<1e-6){ arr[2*i] = island[i][0]; arr[2*i+1] = island[i][0]; }else{ //计算 double x = sqrt(pow(d,2)-pow(island[i][1],2)); arr[2*i] = (double)island[i][0] - x; arr[2*i+1] = (double)island[i][0] + x; } } //排序 for(i=0;i<n-1;i++){ for(j=0;j<n-i-1;j++){ if(arr[2*j]>arr[2*(j+1)]){ double temp = arr[2*(j+1)]; arr[2*(j+1)] = arr[2*j]; arr[2*j] = temp; temp = arr[2*(j+1)+1]; arr[2*(j+1)+1] = arr[2*j+1]; arr[2*j+1] = temp; } } } //去掉重合区间,得到雷达个数 int num = 0; double right = -1; for(i=0;i<n;i++){ if(i==0){ num++; right = arr[2*i+1]; }else{ if(arr[2*i]<=right){ if(arr[2*i+1]<right){ right = arr[2*i+1]; } continue; }else{ num++; right = arr[2*i+1]; } } } return num;}
- poj1328解题报告
- POJ1328解题报告
- poj1328解题报告
- poj1328解题报告.
- POJ1328解题报告
- POJ1328解题报告 (贪心)
- poj1328 Radar Installation 解题报告
- poj1328解题报告(贪心、线段交集)
- POJ1328
- poj1328
- POJ1328
- poj1328
- POJ1328
- poj1328
- poj1328
- poj1328
- poj1328
- poj1328
- SVN服务器搭建和使用(二)
- 从程序员到CTO的Java技术路线图
- 多线程
- Delphi动态创建MainMenu和PopupMenu菜单的方法
- 第六章 Django管理站点
- POJ1328解题报告
- 工作一年之回顾总结展望
- Android 通过按钮Button更改全部的TextView、EditText、Button的字体大小、字体颜色、背景颜色
- 导入android自带的sample源码后gen目录下没有生成R.java
- 清华博士在沪当电力工人 成了拧螺丝最多的人
- Cocos2d-x 中 CCNotificationCenter 的使用
- json解析到listview
- Mocoolka 0.5预览版发布
- ini文件另一种读取办法