12. 17 哈理工网络赛

An Easy Geometry Problem

Description

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

题目分析 : 给你一个所在 n 维空间中的点，会有一些平面， 比如二维空间中的 x = y 平面，现在让你求以所给平面为圆心的半径最大的圆。

首先先考虑二维空间的情况，点到直线的公式这很容易求出来，那么扩展到  n  维空间呢，也是很好想的，任取空间中的两个坐标，投射到相应的平面上，可以计算出一个半径，现在要求所有的半径，直接暴力 n^2 ，肯定会超时，现在你就可以对所给的点进行一个排序，寻找相邻的两个点的最小结果，作为答案。

代码示例 :

　　

#include <cstdio>
#include <iostream>
#include <cmath>
#include <algorithm>
const double num = sqrt(2);
using namespace std;
typedef long long ll;

ll pre[51234];

ll ab(ll x){
	if (x < 0) return -x;
	else return x;
}

int main (){
	int t;
	int n, q;
	int cas = 1;
	
	cin >> t;
	while(t--){
		cin >> n >> q;
		printf("Case #%d:
", cas++);
		while(q--){
			for(int i = 1; i <= n; i++){
				scanf("%lld", &pre[i]);
			}
			sort(pre+1, pre+1+n);  // 这里很重要 
			double ans = 999999999;
			for(int i = 1; i < n; i++){
				ll d = ab(pre[i]-pre[i+1]);
				ans = min(ans, 1.0*d/num);
			}
			printf("%.4lf
", round(ans*10000)/10000);	
		}	
		
		
	}	
	
	return 0;
} 

 I . 

Aggie is faced with a sequence of tasks, each of which with a difficulty value d_i and an expected profit p_i. For each task, Aggie must decide whether or not to complete it. As Aggie doesn’t want to waste her time on easy tasks, once she takes a task with a difficulty d_i, she won’t take any task whose difficulty value is less than or equal to d_i.

Now Aggie needs to know the largest profits that she may gain.

Input

The first line consists of one positive integer t (t ≤ 10), which denotes the number of test cases.

For each test case, the first line consists one positive integer n (n ≤ 100000), which denotes the number of tasks. The second line consists of n positive integers, d₁, d₂, …, d_n (d_i ≤ 100000), which is the difficulty value of each task. The third line consists of n positive integers, p₁, p₂, …, p_n (p_i ≤ 100), which is the profit that each task may bring to Aggie.

Output

For each test case, output a single number which is the largest profits that Aggie may gain.

Sample Input

1

5

3 4 5 1 2

1 1 1 2 2

Sample Output

4

题目分析 ： 类似于LIS，也可以采取维护一个最大的和的序列，但是有一点不同的是，就是在后续插入一个值后，要讲该值后面的元素中键值大于插入的，但是和却小于插入的和的 全部删除掉， 这里用 map 来维护 。

代码示例 ：

int a[eps], b[eps];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t, n;
    map<int, int>mp;
    map<int, int>::iterator it, ip, is;
    
    cin >>t;
    while(t--){
        cin >> n;
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
        mp.clear();
        mp[0] = 0;
        
        int ans = 0;
        for(int i = 1; i <= n; i++){
            if (!mp.count(a[i])) mp[a[i]] = b[i];
            it = mp.find(a[i]);
            ip = --it;
            int p = ip->second + b[i];
            it++;
            mp[a[i]] = max(mp[a[i]], p);
            for(ip = ++it; ip != mp.end(); ip++){
                if (ip->second <= mp[a[i]]) {
                    is = --ip;
                    mp.erase(++ip);
                    ip = is;
                }
                else break;
                
           }
            ans = max(ans, mp[a[i]]);
        }    
        printf("%d
", ans);
    }

    return 0;
}

 解法二 ： 利用 dp 的思想，维护一个最长的递增子序列，将收益视为相同困难度的情况

int a[eps], b[eps];
int dp[eps2];

int main() {
    //freopen("in.txt", "r", stdin);
    //freopen("out.txt", "w", stdout);
    int t, n;
    cin >>t;
    
    while(t--){
        cin >> n;
        for(int i = 1; i <= n; i++) scanf("%d", &a[i]);
        for(int i = 1; i <= n; i++) scanf("%d", &b[i]);
         
        memset(dp, inf, sizeof(dp));
        int len = 0;
        for(int i = 1; i <= n; i++){
            int p = lower_bound(dp, dp+len, a[i])-dp;
            for(int j = p; j < p+b[i]; j++){
                dp[j] = a[i];
            }
            len = max(len, p+b[i]);
        }
        printf("%d
", len);
    }

    return 0;
}