【HDU】1394Minimum Inversion Number-（线段树单点更新，求出逆序数）

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 25122 Accepted Submission(s): 14831

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

Author

CHEN, Gaoli

Source

ZOJ Monthly, January 2003

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代码参考自：https://blog.csdn.net/hellohelloC/article/details/47769657

和：https://blog.csdn.net/u013480600/article/details/22004853

感谢两位大佬！！

代码：

#include<iostream>
#include<cstdio>
#include<cstring>

using namespace std;
const int maxn=50000+5;
int sum[maxn*4];
int a[maxn];
void pushup(int x)
{
    sum[x]=sum[x*2]+sum[x*2+1];
}

void build(int i,int l,int r)
{
    if(l==r)
    {
        scanf("%d",&sum[i]);
        return ;
    }

    int m=(r+l)/2;
    build(i*2,l,m);
    build(i*2+1,m+1,r);
    pushup(i);
}


void update(int p,int i,int l,int r)
{
    if(l==r)
    {
        sum[i]++;
        return ;
    }
    int m=(l+r)/2;
    if(p<=m)update(p,i*2,l,m);
    else update(p,i*2+1,m+1,r);
    pushup(i);
}

int query(int L,int R,int i,int l,int r)
{
    if(L<=l&&r<=R)
    {
        return sum[i];
    }

    int m=(r+l)/2;
    int res=0;
    if(L<=m) res+=query(L,R,i*2,l,m);
    if(m<R) res+=query(L,R,i*2+1,m+1,r);
    return res;
}


int main()
{
    int n;
    while(scanf("%d",&n)==1&&n)
    {
        int ans=0;
        memset(sum,0,sizeof(sum));//也可以在此处选择建树，一样的。。。
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
            ans += query(a[i]+1,n,1,1,n);
            update(a[i],1,1,n);
        }
        int min_num=ans;
        for(int i=1;i<n;i++)
        {
            ans = ans+n+1-2*a[i];
            min_num=min(min_num , ans);
        }
        printf("%d\n",min_num);
    }
    return 0;
}

【HDU】1394Minimum Inversion Number-（线段树单点更新，求出逆序数）

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