Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Problem Description

The inversion number of a given number sequence a1, a2, ..., an 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 will obtain another sequence. There are 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 are 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 two 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 on a single line.

Sample Input

10
1 3 6 9 0 8 5 7 4 2

Sample Output

16

题目大意，从初始的数组开始，每次把第一个加到最后一个，求逆序数，共求出n个逆序数，找出最小值

对每种情况都求逆序数，可以每次都归并排序，或树状数组，或线段树，也可以由上一个的逆序数推出；

a[1] , a[2] , a[3] 。。。a[n]，将a[1]放到最后，然后将a[2]放到最后，可以找到规律，将首个a[i]放到最后时，逆序数增加了 a[i]之前比a[i]大的，增加a[i]之后比a[i]大的，减小了a[i]之前比a[i]小的，减小了a[i]之后比a[i]小的，又因为每次给出的数n个数在0到n，且都不同，最后得出 逆序数会增加 n-a[i]个，减少a[i]-1个
#include <cstdio>
#include <cstring>
#define INF 0x3f3f3f3f
#include <algorithm>
using namespace std;
int tree[100000] , p[6000] , q[6000];
void update(int o,int x,int y,int u)
{
    if( x == y && x == u )
        tree[o]++ ;
    else
    {
        int mid = (x + y)/ 2;
        if( u <= mid )
            update(o*2,x,mid,u);
        else
            update(o*2+1,mid+1,y,u);
        tree[o] = tree[o*2] + tree[o*2+1];
    }
}
int sum(int o,int x,int y,int i,int j)
{
    int ans = 0 ;
    if( i <= x && y <= j )
        return tree[o] ;
    else
    {
        int mid = (x + y) /2 ;
        if( i <= mid )
            ans += sum(o*2,x,mid,i,j);
        if( mid+1 <= j )
            ans += sum(o*2+1,mid+1,y,i,j);
    }
    return ans ;
}
int main()
{
    int i , j , n , min1 , num ;
    while(scanf("%d", &n)!=EOF)
    {
        min1 = 0 ;
        memset(q,0,sizeof(q));
        for(i = 1 ; i <= n ; i++)
        {
            scanf("%d", &p[i]);
            p[i]++ ;
        }
        memset(tree,0,sizeof(tree));
        for(i = 1 ; i <= n ; i++)
        {
            min1 += sum(1,1,n,p[i],n);
            update(1,1,n,p[i]);
        }
        num = min1 ;
        for(i = 1 ; i < n ; i++)
        {
            num = num + ( n - p[i] ) - (p[i] - 1) ;
            if( num < min1 )
                min1 = num ;
        }
        printf("%d\n", min1);
    }
    return 0;
}

hdu1394--Minimum Inversion Number（线段树求逆序数，纯为练习）,布布扣,bubuko.com