hdu 1394 Minimum Inversion Number 线段树 逆序数

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6365    Accepted Submission(s): 3875

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

 

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看不懂怎么证的orz

【

若abcde...的逆序数为k，那么bcde...a的逆序数是多少？我们假设abcde...中小于a的个数为t , 那么大于a的个数就是n-t-1，当把a移动最右位时，原来比a

大的现在都成了a的逆序对，即逆序数增加n-t-1，但是原来比a小的构成逆序对的数，现在都变成了顺序，因此逆序对减少t ,所以新序列的逆序数为 k +=

n - t - t -1，即k += n-1-2 * t , 于是我们只要不断移位（n次），然后更新最小值就可以了

】

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#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#define N 50005
using namespace std;
int num[N];
struct Tree
{
    int l;
    int r;
    int sum;
} tree[N*4];
void build(int root,int l,int r)
{
    tree[root].l=l;
    tree[root].r=r;
    if(tree[root].l==tree[root].r)
    {
        tree[root].sum=0;
        return;
    }
    int mid=(l+r)/2;
    build(root<<1,l,mid);
    build(root<<1|1,mid+1,r);
    tree[root].sum = tree[root<<1].sum+ tree[root<<1|1].sum;
}
void update(int root,int pos,int val)
{
    if(tree[root].l==tree[root].r)
    {
        tree[root].sum=val;
        return;
    }
    int mid=(tree[root].l+ tree[root].r)/2;
    if(pos<=mid)
        update(root<<1,pos,val);
    else
        update(root<<1|1,pos,val);
    tree[root].sum = tree[root<<1].sum+ tree[root<<1|1].sum;
}
int query(int root,int L,int R)
{
    if(L<=tree[root].l&&R>=tree[root].r)
        return tree[root].sum;
    int mid=(tree[root].l+ tree[root].r)/2,ret=0;
    if(L<=mid) ret+=query(root<<1,L,R);
    if(R>mid) ret+=query(root<<1|1,L,R);
    return ret;
}

int x[N];
int main()
{
    int n;
    int sum,ret;
    while (~scanf("%d",&n))
    {
        build(1,0,n-1);
        sum=0;
        for (int i=0;i<n;i++)
        {
            scanf("%d",&x[i]);
            sum+=query(1,x[i],n-1);
            update(1,x[i],1);
        }
        ret=sum;
        for (int i=0;i<n;i++)
        {
            sum+=n-x[i]-x[i]-1;
            if (sum<ret)
            {
                ret=sum;
            }
        }
        printf("%d\n",ret);
    }
    return 0;
}