HDU 1394 tree-like array for reverse order

Minimum Inversion number

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 13036 Accepted Submission (s): 7968

Problem DescriptionThe Inversion number of a given number sequence A1, A2, ..., the number of pairs (AI, aj) that SA Tisfy 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'll Obtain another sequence. There is 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 is asked to write a program to find the minimum inversion number out of the above sequences.

Inputthe input consists of a number of test cases. Each case consists of the 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.

Outputfor each case, output the minimum inversion number to a single line.

Sample Input101 3 6 9 0 8 5 7 4 2

Sample Output16

Authorchen, Gaoli

Sourcezoj Monthly, January 2003 topic meaning: Give an n, then an array of length n, each element is less than n and not duplicated, each operation puts the first element at the end, and the smallest number of reverse order in n operations. Idea: First to find the original array of reverse order number, each time the first place at the end, NUM (reverse order number) plus greater than a[i], and then subtract less than a[i]. Code:

1#include <cstdio>2#include <cstring>3#include <algorithm>4#include <iostream>5#include <vector>6#include <queue>7#include <cmath>8#include <Set>9 using namespacestd;Ten  One #defineN 5005 A #definell Root<<1 - #defineRR root<<1|1 - #defineMid (A[ROOT].L+A[ROOT].R)/2 the  -  - intMaxintXintY) {returnX>y?x:y;} - intMinintXintY) {returnX<y?x:y;} + intAbsintXintY) {returnx<0?-x:x;} -  + intN; A intB[n]; at intA[n]; -  - intLowbit (intx) { -     returnx& (-x); - } -  in voidSolveintValintx) { -      while(x<=N) { toa[x]+=Val; +x+=lowbit (x); -     } the } *  $ intSumintx) {Panax Notoginseng     intans=0; -      while(x>0){ theans+=A[x]; +x-=lowbit (x); A     } the     returnans; + } -  $ Main () $ { -     intT, I, J, K; -      while(SCANF ("%d", &n) = =1){ theMemset (A,0,sizeof(a)); -         intans=0;Wuyi          for(i=1; i<=n;i++) { thescanf"%d",&b[i]); -b[i]++; WuAns+=sum (N)-sum (b[i]); -Solve1, B[i]); About         } $         intnum=ans; -          for(i=1; i<=n;i++){ -num+=n-b[i]-b[i]+1;//sum (n)-sum (B[i])-sum (b[i]-1); -ans=min (ans,num); A         } +printf"%d\n", ans); the     } -}

HDU 1394 tree-like array for reverse order