ACM Minimum Inversion Number Problem solving report-segment tree

C-minimum Inversion numberTime limit:1000MS Memory Limit:32768KB 64bit IO Format:%i64d &%i64 U

Description

The inversion number of a given number sequence A1, A2, ..., 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'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.

Input

The 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.

Output

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

Sample Input

101 3 6 9 0 8 5 7 4 2

Sample Output

16

1 // Segment Tree topic solving.  2// In fact, initially did not understand, and then manually simulated the whole process of building the tree is clear.  3//

1#include"iostream"2#include"algorithm"3#include"Cstdio"4#include"CString"5#include"Cmath"6 #defineMax (A, b) a>b?a:b7 #defineMin (A, b) a<b?a:b8 #defineLson l,m,rt<<19 #defineRson m+1,r,rt<<1|1Ten using namespacestd; One Const intMX =200000+Ten; A intsum[mx<<2]; - voidPushup (intRT) { -sum[rt]=sum[rt<<1]+sum[rt<<1|1];//Update node parent node is the sum of child nodes the } -  - voidBuild (intLintRintRT) { -sum[rt]=0;//Build an empty tree "before this place in the judgment, the leaf node did clear, the side points are missing." +     if(r==l)return ; -     intM= (r+l) >>1; +Build (Lson);//set up left node ABuild (Rson);//Build Right Node at } -  - voidUpdata (intPintLintRintRT) { -     if(r==l) {//find and update target points -sum[rt]++; -         return ; in     } -     intM= (r+l) >>1; to     if(p<=m) Updata (P,lson);//if it's not the target, look around. +     if(P >m) updata (P,rson); -Pushup (RT);//Updates the child nodes for each point that will be updated.  the } *  $ intQuery (intLintRintLintRintRT) {Panax Notoginseng     if(L&LT;=L&AMP;&AMP;R&GT;=R)//size exceeds entire range -         returnSUM[RT];//Total returned the     intM= (r+l) >>1; +     intret=0; A     if(l<= m) ret + = Query (L,r,lson);//the left value of a tree larger than x[i] and the     if(R > m) ret + = Query (L,r,rson);//The right value of a tree larger than x[i] and +     returnret; - } $ intX[MX]; $ intMain () { -     intN; -     intsums; the     Chars[2]; -      while(~SCANF ("%d",&N)) {Wuyisums=0;  theBuild (0, N-1,1);//"It should be better from the 0~n-1, where the location of the 0 from 1~n is not good." The back is just the same. " -          for(intI=0; i<n; i++) { Wuscanf"%d",&x[i]); -Sums+=query (x[i],n-1,0, N-1,1); AboutUpdata (X[i],0, N-1,1); $         } -         intret=sums; -          for(intI=0; i<n; i++) { -sums=sums+n-2*x[i]-1; Aret=min (ret,sums); +         } theprintf"%d\n", ret); -     } $     return 0; the}

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ACM Minimum Inversion Number Problem solving report-segment tree