Minimum Inversion Number

Minimum Inversion Number

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status

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

 

 

 

#include <stdio.h>
#include <algorithm>

using namespace std;

int a[5005];
struct Node{
    int l,r,num;
}tree[50000];

void Build(int n,int x,int y){
    tree[n].l = x;
    tree[n].r = y;
    tree[n].num = 0;
    if(x == y){
        return;
    }
    int mid = (x + y) / 2;
    Build(2*n,x,mid);
    Build(2*n+1,mid+1,y);
}

void Modify(int n,int x){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    if(x == l && x == r){
        tree[n].num = 1;
        return;
    }
    if(x <= mid)    Modify(2*n,x);
    else            Modify(2*n+1,x);
    tree[n].num = tree[2*n].num + tree[2*n+1].num;
}

int Query(int n,int x,int y){
    int l = tree[n].l;
    int r = tree[n].r;
    int mid = (l + r) / 2;
    int ans = 0;;
    if(x == l && y == r)
        return tree[n].num;
    if(x <= mid)   ans += Query(2*n,x,min(mid,y));
    if(y > mid)    ans += Query(2*n+1,max(mid+1,x),y);
    return ans;
}
int main(){
    //freopen ("a.txt" , "r" , stdin ) ;
    int n,sum,ans;
    int i,j;

    while(scanf("%d",&n) != EOF){
        sum = 0;
        Build(1,0,n);
        for(i = 1;i <= n;i++){
            scanf("%d",&a[i]);
            Modify(1,a[i]);
            sum += Query(1,a[i]+1,n);
        }
        ans = sum;
        for(i = 1;i < n;i++){
            sum = sum + (n - 1 - a[i]) - a[i];
            if(sum < ans)
                ans = sum;
        }
        printf("%d\n",ans);
    }
}