Minimum Inversion Number_线段树||树状数组

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

 

【题意】给出n个数，求其在第一个数到最后一个的过程中，最小的逆序数是多少

【思路】建立一颗空树，在插入每个数的时候，统计这个数前面有多少数大于他，当a[i]由第一个变为最后一个时，要加上a[i]后面大于a[i]的数的个数，有n-1-a[i]个，要减去a[i]后面小于a[i]的数的个数，有a[i]个（注意i是从0开始的）

#include<iostream>
#include<string.h>
#include<stdio.h>
using namespace std;
const int N=5005;
struct node
{
    int l,r;
    int num;

}tree[N*4];
void build(int k,int l,int r)//建空树
{
    int mid=(l+r)/2;
    tree[k].l=l;
    tree[k].r=r;
    tree[k].num=0;
    if(l==r) return ;
    build(k*2,l,mid);
    build(k*2+1,mid+1,r);
}
void updata(int k,int c)//插入
{
    if(tree[k].l==c&&tree[k].r==c)
    {
        tree[k].num=1;
        return ;
    }
    int mid=(tree[k].l+tree[k].r)/2;
    if(c<=mid)
        updata(k*2,c);
    else updata(k*2+1,c);
    tree[k].num=tree[k*2].num+tree[k*2+1].num;
}
int get_sum(int k,int c,int n)//统计
{
    if(c<=tree[k].l&&tree[k].r<=n) return tree[k].num;
    else
    {
        int mid=(tree[k].l+tree[k].r)/2;
        int sum1=0,sum2=0;
        if(c<=mid)
            sum1=get_sum(k*2,c,n);
        if(n>mid)
            sum2=get_sum(k*2+1,c,n);
        return sum1+sum2;
    }
}


int main()
{
    int n;
    while(scanf("%d",&n)>0)
    {
        int a[N];
        build(1,0,n-1);
        int ans=0;
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            ans+=get_sum(1,a[i]+1,n-1);
//每输入一个数时，检验一下先他输入的并比他大的数的个数，对于前面比他大的数来说，他是他们的逆序数
            updata(1,a[i]);
        }
        int minx=ans;
        for(int i=0;i<n;i++)
        {
            ans=ans+n-2*a[i]-1;           //当a[i]由第一个变为最后一个时，要加上a[i]后面大于a[i]的数的个数，有n-1-a[i]个，要
            if(ans<minx)                    //减去a[i]后面小于a[i]的数的个数，有a[i]个（注意i是从0开始的）
                minx=ans;
        }
        printf("%d
",minx);
    }
    return 0;
}

#include<iostream>
#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
const int N=100000+10;
int a[N],b[N],c[N],ans[N];
int n,cnt,sum;
int lowbit(int x)
{
    return x&(-x);
}
int query(int x)
{
    int res=0;
    x++;
    while(x<=n)
    {
        res+=c[x];
        x+=lowbit(x);
    }
    return res;
}
void update(int x)
{
    while(x)//对在他前且比他小的数，逆序数加1；
    {
        c[x]++;
        x-=lowbit(x);
    }
}
int main()
{
    while(~scanf("%d",&n))
    {
        memset(a,0,sizeof(a));
        memset(c,0,sizeof(c));
        int sum=0;
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            update(a[i]+1);
            sum+=query(a[i]+1);
        }
        int tmp=0,s=sum;
        for(int i=1;i<=n;i++)
        {
            tmp=sum-a[i]+(n-a[i]-1);
            sum=tmp;
            if(tmp<s)
                s=tmp;
        }
        printf("%d
",s);
    }
    return 0;
}