HDU 1394 Minimum Inversion Number

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6471    Accepted Submission(s): 3940

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

求最小逆序数，n个数，有n种形式

线段树和树状数组都能做。

线段树代码

#include<stdio.h>

struct st

{

    int l;

    int r;

    int sum;

}f[4*5002];

int date[5002];

void build(int l,int r,int n)

{

    int mid=(l+r)/2;

    f[n].l=l;

    f[n].r=r;

    if(l==r)

        f[n].sum=date[l];

    else

    {

        build(l,mid,n*2);

        build(mid+1,r,n*2+1);

        f[n].sum=f[n*2].sum+f[n*2+1].sum;

    }

}

void update(int wz,int num,int n)

{

    int mid=(f[n].l+f[n].r)/2,t;

    if(f[n].l==wz&&f[n].r==wz)

    {

        f[n].sum=f[n].sum+num;

        t=n/2;

        while(t)

        {

            f[t].sum=f[t*2].sum+f[t*2+1].sum;

            t=t/2;

        }

    }

    else if(mid>=wz)

        update(wz,num,n*2);

    else update(wz,num,n*2+1);

}

int getsum(int l,int r,int n)

{

    int mid=(f[n].l+f[n].r)/2;

    if(f[n].l==l&&f[n].r==r)

        return f[n].sum;

    else if(mid>=r)

        getsum(l,r,n*2);

    else if(mid<l)

        getsum(l,r,n*2+1);

    else 

    {

        return getsum(l,mid,n*2)+getsum(mid+1,r,n*2+1);

    }

}

int main()

{

    int i,k,n,sum,s;

    while(scanf("%d",&n)>0)

    {

        for(i=1;i<=n;i++)

            date[i]=0;

        build(1,n,1);

        sum=0;

        for(i=1;i<=n;i++)

        {

            scanf("%d",&date[i]);

            date[i]++;

        }

        for(i=1;i<=n;i++)

        {

            update(date[i],1,1);

            k=getsum(1,date[i],1);

            sum=sum+i-k;

        }

        s=sum;

        for(i=1;i<=n;i++)

        {

        //    printf("%d ",s);

            k=n-date[i]-(date[i]-1);

            s=s+k;

            if(s<sum)

                sum=s;    

        }

        printf("%d\n",sum);

    }

    return 0;

}

 

树状数组代码：

#include<stdio.h>

int a[5003],d[5003];

int lowbit(int x)

{

    return x&(-x);

}

int sum(int x,int *p)

{

    int k=0;

    while(x)

    {

        k=k+p[x];

        x=x-lowbit(x);

    }

    return k;

}

void add(int x,int n,int num,int *p)

{

    int i;

    for(i=x;i<=n;i=i+lowbit(i))

        p[i]=p[i]+num;

}

int main()

{

    int i,j,k,n,m,cz,k1,k2,min;

    while(scanf("%d",&n)>0)

    {

        for(i=1;i<=n;i++)

        {

            scanf("%d",&a[i]);

            a[i]++;

            d[i]=0;

        }

        for(i=1,cz=0;i<=n;i++)

        {

            add(a[i],n,1,d);

            k1=i-sum(a[i],d);

            cz=k1+cz;

        }

    //    printf("%d\n",cz);

        for(i=1,k1=0,min=cz;i<n;i++)

        {

            k1=n-a[i];

            k2=-a[i]+1;

            min=k1+k2+min;

        //    printf("%d ",k1+k2);

            if(min<cz)

                cz=min;

        }

        printf("%d\n",cz);

    }

    return 0;

}