ZYB's Premutation(挺简单的数据结构题目)

本文链接：https://blog.youkuaiyun.com/ZSGG_ACM/article/details/50210423

本文探讨了一个关于序列重构的问题，通过分析给定的前缀逆序对数量，求解原始序列。采用二分搜索与区间树状数组进行优化，有效解决了复杂度较高的输入数据处理。

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ZYB's Premutation

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 554 Accepted Submission(s): 271

Problem Description

ZYB has a premutation

P ,but he only remeber the reverse log of each prefix of the premutation,now he ask you to
restore the premutation.

Pair

(i,j)(i<j) is considered as a reverse log if

Ai>Aj is matched.

Input

In the first line there is the number of testcases T.

For each teatcase:

In the first line there is one number

N .

In the next line there are

N numbers

Ai ,describe the number of the reverse logs of each prefix,

The input is correct.

1≤T≤5 ,

1≤N≤50000

Output

For each testcase,print the ans.

Sample Input

Sample Output

  
   3 1 2

Source

BestCoder Round #65

题意： HDU上有中文题面，题意我就不呵呵了

题解：很容易想到，从数组后面往前推，因为最后一个数我们可以确定其位置，接下来就是代码优化问题。直接暴力跑的话是会超时的。这题我用的是二分位置，加区间树状数组验证答案。

AC代码：

import java.util.*;
import java.io.*;

public class main{
	
	static Scanner cin = new Scanner(System.in);
	static PrintWriter out = new PrintWriter(System.out);
	static int[] a = new int[50005];
	static int[] c = new int[50005];
	static int n;
	
	public static void main(String[] args) throws IOException{
		int T = cin.nextInt();
		while(T-- > 0) {
			int pre = -1,now;
			n = cin.nextInt();
			for(int i = 1; i <= n; i++) {
				a[i] = cin.nextInt();
				if(pre == -1) pre = a[i];
				else {
					now = a[i] - pre; 
					pre = a[i];
					a[i] = now; 
				}
			}
			Arrays.fill(c, 0);
			for(int i = n; i >= 1; i--) {
				int l = 1,r = n,ans = 0;
				while(l <= r) {
					int mid = l + r >> 1;
					int tp = n - mid - que(mid + 1, n);
					if(tp > a[i] || tp == a[i] && que(mid, mid) == 0) {
						ans = mid;
						l = mid + 1;
					} else {
						r = mid - 1;
					}
				}
				a[i] = ans;
				add(ans);
			}
			boolean ok = false;
			for(int i = 1; i <= n; i++) {
				if(ok) System.out.print(' ');
				else ok = true;
				System.out.print(a[i]);
			}
			System.out.println();
		}
	}
	public static void add(int x) {
		while(x <= n) {
			c[x]++;
			x += x & -x;
		}
	}
	public static int que(int l,int r) {
		int s = 0;
		if(l > r) return s;
		while(r > 0) {
			s += c[r];
			r -= r & -r;
		}
		l--;
		while(l > 0) {
			s -= c[l];
			l -= l & -l;
		}
		return s;
	}
}