hdu1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 8317    Accepted Submission(s): 5109

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
   
   
   
   

 

Author

CHEN, Gaoli

 

Source

ZOJ Monthly, January 2003

 

Recommend

Ignatius.L

 

线段树，第一遍初始化的时候计算出逆序对，然后之后每次把第一个数移动到最后一个位置的时候需要把比他小的那部分数都减一，然后这个数的值就应该是比他大的数的个数，因为，移到最后一个位置上了所有数都在他前面。。。。

代码

#include <stdio.h>
#include <string.h>
#include <algorithm>
using namespace std;

typedef struct
{
    int l,r,val,lazy,show;
}Tree;

Tree tree[50000];
int a[5005];

void Push(int t)
{
    tree[2*t+1].val+=(tree[2*t+1].r-tree[2*t+1].l+1)*tree[t].lazy;
    tree[2*t+2].val+=(tree[2*t+2].r-tree[2*t+2].l+1)*tree[t].lazy;
    tree[2*t+1].lazy+=tree[t].lazy;
    tree[2*t+2].lazy+=tree[t].lazy;
    tree[t].lazy=0;
}

void Build(int t,int l,int r)
{
    tree[t].l=l;
    tree[t].r=r;
    tree[t].val=0;
    tree[t].lazy=0;
    tree[t].show=0;
    if (l==r) return;
    int mid=(l+r)>>1;
    Build(2*t+1,l,mid);
    Build(2*t+2,mid+1,r);
}

void point_update(int t,int x,int val)
{
    if (tree[t].l==tree[t].r)
    {
        tree[t].val=val;
        tree[t].show=1;
        return;
    }
    if (tree[t].lazy!=0)
    {
        Push(t);
    }
    int mid=(tree[t].l+tree[t].r)>>1;
    if (x<=mid) point_update(2*t+1,x,val);
    else point_update(2*t+2,x,val);
    tree[t].val=tree[2*t+1].val+tree[2*t+2].val;
    tree[t].show=tree[2*t+1].show+tree[2*t+2].show;
}

void line_update(int t,int l,int r,int val)
{
    if (l>r) return;
    if (tree[t].l==l && tree[t].r==r)
    {
        tree[t].val+=(tree[t].r-tree[t].l+1)*val;
        tree[t].lazy+=val;
        return;
    }
    if (tree[t].lazy!=0)
    {
        Push(t);
    }
    int mid=(tree[t].l+tree[t].r)>>1;
    if (l<=mid) line_update(2*t+1,l,min(mid,r),val);
    if (r>mid) line_update(2*t+2,max(l,mid+1),r,val);
    tree[t].val=tree[2*t+1].val+tree[2*t+2].val;
}

int query_show(int t,int l,int r)
{
    if (tree[t].l==l && tree[t].r==r)
    {
        return tree[t].show;
    }
    int mid=(tree[t].l+tree[t].r)>>1;
    int ret=0;
    if (l<=mid) ret+=query_show(2*t+1,l,min(mid,r));
    if (mid<r) ret+=query_show(2*t+2,max(mid+1,l),r);
    return ret;
}

int main()
{
    int i,j,n,ans;
    while(scanf("%d",&n)!=EOF)
    {
        Build(0,0,n-1);
        for (i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            point_update(0,a[i],query_show(0,a[i],n-1));
        }
        ans=tree[0].val;
        for (i=0;i<n;i++)
        {
            point_update(0,a[i],n-a[i]-1);
            line_update(0,0,a[i]-1,-1);
            ans=min(ans,tree[0].val);
        }
        printf("%d\n",ans);
    }
    return 0;
}