ACM: 序列中找最小逆序对 杂题 TOJ…

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本文介绍了一种算法，用于解决给定序列通过特定变换后得到的序列中逆序对数量最小的问题。该算法首先计算初始序列的逆序对数量，然后通过分析序列变换前后逆序对数量的变化规律，逐步迭代计算所有可能序列的逆序对数量，并最终确定出逆序对数量最小的序列。

Minimum Inversion Number

描述

The inversion number of a given number sequence a1, a2, ..., an is the number of pairs (ai, aj) that satisfy i < j and ai > aj.

For a given sequence of numbers a1, a2, ..., an, if we move the first m >= 0 numbers to the end of the seqence, we will obtain another sequence. There are totally n such sequences as the following:

a1, a2, ..., an-1, an (where m = 0 - the initial seqence)
a2, a3, ..., an, a1 (where m = 1)
a3, a4, ..., an, a1, a2 (where m = 2)
...
an, a1, a2, ..., an-1 (where m = n-1)

You are asked to write a program to find the minimum inversion number out of the above sequences.

输入

The input consists of a number of test cases. Each case consists of two lines: the first line contains a positive integer n (n <= 5000); the next line contains a permutation of the n integers from 0 to n-1.

输出

For each case, output the minimum inversion number on a single line.

样例输入

1 3 6 9 0 8 5 7 4 2

样例输出

题意: 在规定的变化序列中找出最少逆序对的个数.

解题思路:

1. 先计算出初始序列的逆序对的个数.

2. 再计算把当前序列最后一个数放到最前面的时候逆序对变化.

问题分析: 假设 sum 表示当前序列的逆序对数. num表示当前序列最后一个数.

要计算变化的情况首先, 假设最坏的情况. 即: sum -= (n-1);

再 sum += (num的逆序对)*2

3. 不断更新最小值.

代码:

#include <cstdio>
#include <iostream>
#include <cstring>
using namespace std;
#define MAX 5005

int n;
int sum;
int minsize;
int a[MAX];

int main()
{
freopen("input.txt","r",stdin);
int i, j;
while(scanf("%d",&n) != EOF)
{
  sum = 0;
  for(i = 0; i < n; ++i)
   scanf("%d",&a[i]);

  for(i = 0; i < n; ++i)
  {
   for(j = i+1; j < n; ++j)
   {
    if(a[i] > a[j])
     sum++;
   }
  }

  minsize = sum;
  for(i = n-1; i > 0; --i)
  {
   sum = (sum - (n-1));
   sum = (sum + (a[i]*2));
   if(minsize > sum)
    minsize = sum;
  }
  printf("%d\n",minsize);
}
return 0;
}