HDU 1394Minimum inversion number array in reverse order to number and

Minimum Inversion number

Time limit:2000/1000 MS (java/others) Memory limit:65536/32768 K (java/others)
Total submission (s): 18543 Accepted Submission (s): 11246

problem DescriptionThe 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.

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 in a single line.

Sample Input101 3 6 9 0 8 5 7 4 2

Sample Output16

Authorchen, Gaoli

Source ZOJ Monthly, January 2003

RECOMMENDIGNATIUS.L | We have carefully selected several similar problems for you:1166 1698 1540 1542 1255title Link: http://acm.hdu.edu.cn/showproblem.php?pid=1394 Test Instructions: Find out the total number of pairs (Ai,aj) (I<j,ai>aj). That is, the number of the AI on the right is smaller than AI. Each time the sequence of a is transformed, the minimum number of inverse pairs is calculated. idea: Because the most basic function of a tree array is to find the number of points smaller than a point x . So reverse store AI. Code:

#include <iostream>#include<cstdio>#include<cstring>using namespacestd;Const intmaxn=1e6+ -, inf=1e9+ -;intA[MAXN],C[MAXN],ANS[MAXN];intLowbit (intx) {    returnx& (-x);}voidAddintIintval) {     for(i; i<=maxn; i+=lowbit (i)) C[i]+=Val;}intSuminti) {    ints=0;  for(i; i>0; i-=lowbit (i)) s+=C[i]; returns;}intMain () {intI,j,t,n;  while(SCANF ("%d", &n)! =EOF) {         for(I=n; i>=1; i--) scanf ("%d",&A[i]); Memset (c,0,sizeof(c)); memset (ans,0,sizeof(ans)); intcou=0;  for(i=1; i<=n; i++) {Ans[i]=sum (a[i]+1); Cou+=Ans[i]; Add (A[i]+1,1); }        intmin=cou;  for(I=n; i>1; i--)        {             for(j=1; j<=n; J + +)            {                if(j==i)Continue; if(a[i]<A[j]) {Cou++; ANS[J]++; }} cou-=Ans[i]; Ans[i]=0; if(cou<min) min=cou; } cout<<Min<<Endl; }    return 0;}

reverse order to

HDU 1394Minimum inversion number array in reverse order to number and