hrbust 2354 An Easy Geometry Problem

An Easy Geometry ProblemTime Limit: 1000 MSMemory Limit: 262144 KTotal Submit: 52(25 users)Total Accepted: 19(16 users)Rating: Special Judge: NoDescription

Lily is interested in hyperspace recently, and she is trying to solve an easy problem now. Given an n-dimensional hyperspace, assuming each dimension is a₁, a₂, a₃, ..., a_n. And for each i (1 ≤ i ≤ n), j (i < j ≤ n), there is a plane a_i=a_j  These planes have divided the hyperspace into different regions. For example, in 3D hyperspace, we have three planes: x=y, y=z and x=z.

Now given q queries, for each query, there is a point (x₁, x₂, x₃, ..., x_n). Lily wants to know the maximum radius r of an n-dimensional ball she can put in point (x₁, x₂, x₃, ..., x_n), which means the center of the ball is in point (x₁, x₂, x₃, ..., x_n) and the ball’s surface cannot cross but can touch any planes mentioned above.

The equation of an n-dimensional ball with center (x₁, x₂, x₃, ..., x_n) is . And the distance between two points (x₁, x₂, x₃, ..., x_n) and (x₁', x₂', x₃', ..., x_n') in an n-dimensional hyperspace is:.

Input

First line contains an integer T (1 ≤ T ≤ 5), represents there are T test cases.

For each test case: First line contains two integers n (3 ≤ n ≤ 50000) and q (1 ≤ q ≤ 8), where n is the dimension of the hyperspace, and q is the number of queries. Following q lines, each line contains n integers x₁, x₂, x₃, ..., x_n (-10⁹ ≤ x_i ≤ 10⁹), represents the coordinate of the ball’s center.

Output

For each test case, output one line containing "Case #X:"(without quotes), where X is the case number starting from 1, and output q lines containing the maximum possible radius of the ball in each query in input order, rounded to four decimal places.

Sample Input

1

3 2

0 0 0

11 22 33

Sample Output

Case #1:

0.0000

7.7782

Source“科林明伦杯”哈尔滨理工大学第七届程序设计团队赛
读题两小时实现5分钟。

最大圆半径。（满足_j   These planes have divided the hyperspace into different regions. For example, in 3D hyperspace, we have three planes: x=y, y=z and x=z ）

转化成2维直接找到规律。

#include<stdio.h>#include<math.h>#include<algorithm>using namespace std;#define N 100005int T;double r[N];int main(){    scanf("%d",&T);    int icase=0;    while(T--){        int n,m;        printf("Case #%d:\n",++icase);        scanf("%d%d",&n,&m);        while(m--){            for(int i=0;i<n;i++){                scanf("%lf",&r[i]);            }            sort(r,r+n);            double res = 1e15;            for(int i=1;i<n;i++){                res = min(res,r[i]-r[i-1]);            }            printf("%.4f\n",res/sqrt(2.0));        }    }}