HDU 1394 Minimum Inversion number

Minimum Inversion number

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

1#include <cstdio>2#include <cstring>3#include <algorithm>4 using namespacestd;5 6 structSegtree7 {8     intL,r;9     intnum;Ten     intMid () One     { A         return(l+r) >>1; -     } - }; the  -Segtree tree[5005<<2]; - inta[5005<<2]; -  + voidPushup (intRT) - { +tree[rt].num=tree[rt<<1].num+tree[rt<<1|1].num; A } at  - voidBuildintLintRintRT) - { -Tree[rt].l=l; -Tree[rt].r=R; -tree[rt].num=0; in     if(L==R)return; -     intm=Tree[rt].mid (); toBuild (l,m,rt<<1); +Build (m+1,r,rt<<1|1); - } the  * intQueryintLintRintRT) $ {Panax Notoginseng     if(tree[rt].l==l&&tree[rt].r==R) -         returnTree[rt].num; the     intm=Tree[rt].mid (); +     if(r<=m) A         returnQuery (l,r,rt<<1); the     Else if(l>m) +         returnQuery (l,r,rt<<1|1); -     Else $         returnQuery (l,m,rt<<1) +query (m+1,r,rt<<1|1); $ } -  - voidUpdateintPosintRT) the { -     if(tree[rt].l==TREE[RT].R)Wuyi     { thetree[rt].num++; -         return; Wu     } -     intm=Tree[rt].mid (); About     if(pos<=m) Update (pos,rt<<1); $     ElseUpdate (pos,rt<<1|1); - pushup (RT); - } -  A intMain () + { the     intn,i,j; -      while(SCANF ("%d", &n)! =EOF) $     { the          for(i=0; i<n;i++) thescanf"%d",&a[i]); theBuild0, N-1,1); the         intsum=0; -          for(i=0; i<n;i++) in         { theSum+=query (a[i],n-1,1); theUpdate (A[i],1); About         } the         intans=999999999; theans=min (ans,sum); the          for(i=0; i<n;i++) +         { -sum=sum-a[i]+n-1-A[i]; theans=min (ans,sum);Bayi         } theprintf"%d\n", ans); the     } -     return 0; -}

HDU 1394 Minimum Inversion number