HDU 1394- Minimum Inversion Number(线段树求逆序数)

Minimum Inversion Number

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

Submit Status Practice HDU
1394

Appoint description: 
System Crawler  (2015-04-13)

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：求序列的逆序数，每求完一次将该序列的第一个数放在最后一个，然后再求逆序数，一直重复，然后求最小的逆序数。

思路：用线段树记录下各个数，区间的值[a，b]表示数字a～b的已经出现了多少次，所以对于ai，只需要查询[ai, n]有多少个(ai之后的比ai小的有多少个)，就是代表ai有多少个逆序数了。然后通过推理我们可以发现：假如目前的第一个数是a[i]，那当把他放到最后面的时候，少的逆序数是本来后面比他小的数的个数。多的逆序数就是放到后面后前面比他大的数的个数。因为所有数都是从0到n-1.所以比他小的数就是a[i]，比他大的数就是n-1-a[i]。所以算出初始序列的逆序数，然后用出事序列的逆序数+n-2*a[i]-1就是改变后的序列的逆序数。

#include <stdio.h>
#include <math.h>
#include <string.h>
#include <stdlib.h>
#include <iostream>
#include <sstream>
#include <algorithm>
#include <set>
#include <queue>
#include <stack>
#include <map>
using namespace std;
typedef long long LL;
const int inf=0x3f3f3f3f;
const double pi= acos(-1.0);
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
const int MAXN=5010;
int sum[MAXN<<2];

void PushUp(int rt)
{
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}

void Build(int l,int r,int rt)
{
    sum[rt]=0;
    if(l==r) {
        return ;
    }
    int mid=(l+r)>>1;
    Build(lson);
    Build(rson);
    PushUp(rt);
}

void Update(int p,int l,int r,int rt)
{
    if(l==r) {
        sum[rt]++;
        return ;
    }
    int mid=(l+r)>>1;
    if(p<=mid)
        Update(p,lson);
    else
        Update(p,rson);
    PushUp(rt);
}

int Query(int ll,int rr,int l,int r,int rt)
{
    if(ll<=l&&rr>=r) {
        return sum[rt];
    }
    int mid=(l+r)/2;
    int ans=0;
    if(ll<=mid)
        ans+=Query(ll,rr,lson);
    if(rr>mid)
        ans+=Query(ll,rr,rson);
    return ans;
}

int main()
{
    int n,i;
    int a[MAXN];
    int ans;
    int x;
    while(~scanf("%d",&n)) {
        ans=0;
        Build(0,n-1,1);
        for(i=0; i<n; i++) {
            scanf("%d",&a[i]);
            ans+=Query(a[i],n-1,0,n-1,1);
            Update(a[i],0,n-1,1);
        }
        x=ans;
        for(i=0; i<n; i++) {
            x+=n-2*a[i]-1;
            ans=min(ans,x);
        }
        printf("%d\n",ans);
    }
    return 0;
}