POJ 2299 ultra-quicksort Merge sort to find the inverse number pair

Topic Links:

http://poj.org/problem?id=2299

Title Description:

To a chaotic sequence with n (n<=500000), Q: If you sort the n into ascending order, how many times do you need to exchange it?

Problem Solving Ideas:

According to the characteristics of bubble sort, we know that the problem only need to count each number of reverse order number (if there is i<j, existence a[i] > A[j], then called A[i] and

A[J] for reverse number pairs), the output of all the number of reverse-order number and the normal sort must be timed out, but the relatively fast sort, like fast row and can not be counted

The number of exchanges, here is a good way to reflect the advantages of the merger sort. The typical use of merge sort to find the number of reverse order.

Merge sort: For example now there is a sequence [l,r], we can divide this sequence into two sequences [L,mid], [mid,r], using recursion to follow the

The method is gradually reduced Sequence, the order of the sub-sequence, and then order the sub-sequence interval, and then merge the ordered interval, a good drop embodies the idea of division.

Code:

1#include <cstdio>2#include <cstring>3#include <cstdlib>4#include <iostream>5 using namespacestd;6 #defineMAXN 5000107 8 intA[MAXN], B[MAXN];9 Long Longcount;Ten  One voidMerge (intLintR); A intMain () - { -     intn, I; the      while(SCANF ("%d", &N), N) -     { -Memset (A,0,sizeof(a)); -memset (b,0,sizeof(b)); +          for(i=0; i<n; i++) -scanf ("%d", &a[i]); +Count =0;//must be used int64,int32 will overflow AMerge (0, n); atprintf ("%lld\n", count); -     } -     return 0; - } -  - voidMerge (intLintr)//merge sort, the parameter is the position of the sub-interval respectively in { -     if(R-l <=1) to         return ; +     intMID = L + (r-l)/2; - merge (L, mid); the Merge (Mid, R); *     intx = L, y = Mid, I =l; $      while(X<mid | | y<r)//order the subsequence and save in array bPanax Notoginseng     { -         if(Y >= R | | (X < mid && a[x] <=A[y])) theB[i + +] = a[x + +]; +         Else A         { the             if(X <mid)//record number of exchanges +Count + = mid-x; -B[i + +] = A[y + +]; $         } $     } -      for(i=l; i<r; i++)//Copy the ordered sub-sequence to the corresponding position of the A array -A[i] =B[i]; the}

POJ 2299 ultra-quicksort Merge sort to find the inverse number pair