题目描述

Let { A1,A2,…,An } be a permutation of the set{ 1,2,…, n}. If i < j and Ai > Aj then the pair (Ai,Aj) is called an “inversion” of the permutation. For example, the permutation {3, 1, 4, 2} has three inversions: (3,1), (3,2) and (4,2).
The inversion table B1,B2,…,Bn of the permutation { A1,A2,…,An } is obtained by letting Bj be the number of elements to the left of j that are greater than j. (In other words, Bj is the number of inversions whose second component is j.) For example, the permutation:
{ 5,9,1,8,2,6,4,7,3 }
has the inversion table
2 3 6 4 0 2 2 1 0
since there are 2 numbers, 5 and 9, to the left of 1; 3 numbers, 5, 9 and 8, to the left of 2; etc.
Perhaps the most important fact about inversions is Marshall Hall’s observation that an inversion table uniquely determines the corresponding permutation. So your task is to convert a permutation to its inversion table, or vise versa, to convert from an inversion table to the corresponding permutation.

Input:

The input consists of several test cases. Each test case contains two lines.
The first line contains a single integer N ( 1 <= N <= 50) which indicates the number of elements in the permutation/invertion table.
The second line begins with a single charactor either ‘P’, meaning that the next N integers form a permutation, or ‘I’, meaning that the next N integers form an inversion table.
Following are N integers, separated by spaces. The input is terminated by a line contains N=0.

Output:

For each case of the input output a line of intergers, seperated by a single space (no space at the end of the line). If the input is a permutation, your output will be the corresponding inversion table; if the input is an inversion table, your output will be the corresponding permutation.

Sample Input:

9
P 5 9 1 8 2 6 4 7 3
9
I 2 3 6 4 0 2 2 1 0
0

Sample Output:

2 3 6 4 0 2 2 1 0
5 9 1 8 2 6 4 7 3

核心思路

• permutation -> inversion:
  因为n的数据范围不大，所以可以用朴素版的双指针算法。
  双指针 ： i 指向目标数字 ，j 从头向后扫描寻找比i要大的数，遇到i 就回滚，寻找一个数 i+1；

• inversion -> permutation:
  类似哈希表的方法。

附带本人提交情况

提交运行时间，4毫秒。

C++代码

#include<iostream> 
#include<cstring> 
using namespace std;

const int N = 60;
int a[N],t[N];
int n;
char c;

void permutation_to_inversion()
{
    
    
	for(int i=1;i<=n;i++)          // 双指针 ：i 指向目标数字 ，j 从头向后扫描寻找比i要大的数，遇到i 就回滚，寻找一个数 i+1； 
	{
    
      int j=1;
		while(j<=n)
		{
    
    
   	 	if(a[j]==i)break;
			if(a[j]>i)
			{
    
    
				t[i]++; j++;
        	}
        	else j++; 
   	 
       }
    }
}

void inversion_to_permutation()
{
    
    
	int i,j,count,k;
   for(j=1; j<=n; j++)
   {
    
    
       count = 0;
    
       for(i=1; i<=n && count<a[j]; i++)
       {
    
                                                             // 扫描寻找 符合的位置
           if( t[i]==0 )
               count++;
       }
      
       if( t[i] != 0)     // 位置发生了冲突。
           for(k=i+1; k<=n; k++)
               if( t[k] == 0 ) 
               {
    
    
                   i = k;
                   break;
               }
       t[i] = j;      
   }
  
}

int main()
{
    
    
	ios::sync_with_stdio(false);
    cin.tie(nullptr);
   while(cin>>n&&n!=0)
	{
    
    

		cin>>c;
	
	    memset(a,0,sizeof(a));
	    memset(t,0,sizeof(t));
	    
   	for(int i=1;i<=n;i++) cin>>a[i];
		
   	if(c=='P') permutation_to_inversion();
       else inversion_to_permutation();
  		
   	for(int i=1;i<=n;i++) 
   	{
    
       if(i!=1)cout<<" ";
   		cout<<t[i];
       }
       cout<<"\n";
   
   }
	return 0;
}