线段树单点更新 HDU 1394 Minimum Inversion Number

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：求最小逆序数。

分析：线段树的一个应用，求逆序数。

从a[0]开始扫描，找比a[0]大的个数，也就是询问区间(a[0],n-1)中点的个数，然后再插入a[0]，更新父节点即可。

思路为：tree[i]为0代表i未出现，为1则已经出现，
然后查询时我们只要统计新加入的数到n-1这个范围内有多少个tree[i]为1，也就是当前
数的逆序，然后累加，就是我们要求的逆序数。因为其他形式都可以由第一次求得的逆
序数递推而来，所以只需计算一次。

有个结论，设逆序数为sum，x[0]后面比它小的一定是x[0]个。那么移到末尾后，比x[0]大的数的后面比它小的数统统加一，也就是加（n - a[0] - 1），然后它放到末尾了，他原来的后面比它小的数变为0，也就是sum = sum + (n - a[0] - 1) - a[0];

AC代码：

#include<stdio.h>
#define maxn 5010
#define lson l,m,rt << 1
#define rson m+1,r,rt << 1 | 1
int tree[maxn<<2],x[maxn];
int min(int a,int b)
{
    return a<b?a:b;
}
void PushUP(int rt)
{
    tree[rt]=tree[rt<<1]+tree[rt<<1|1];
}
void build(int l,int r,int rt)
{
    tree[rt]=0;
    if(l==r)
        return ;
    int m=(l+r)>>1;
    build(lson);
    build(rson);
}
int query(int L,int R,int l,int r,int rt)
{
    if(L<=l&&R>=r)
        return tree[rt];
    int m=(l+r)>>1;
    int ret=0;
    if(L<=m)
        ret+=query(L,R,lson);
    if(R>m)
        ret+=query(L,R,rson);
    return ret;
}
void update(int p,int l,int r,int rt)
{
    if(l==r)
    {
        tree[rt]++;
        return ;
    }
    int m=(l+r)>>1;
    if(p<=m)
        update(p,lson);
    else
        update(p,rson);
    PushUP(rt);
}
int main()
{
    int i,n,sum;
    while(scanf("%d",&n)!=EOF)
    {
        build(0,n-1,1);
        sum=0;
        for(i=0;i<n;i++)
        {
            scanf("%d",&x[i]);
            sum+=query(x[i],n-1,0,n-1,1);
            update(x[i],0,n-1,1);
        }
        int ret=sum;
        for(i=0;i<n;i++)
        {
            sum+=n-x[i]-1-x[i];
            ret=min(ret,sum);
        }
        printf("%d\n",ret);
    }
    return 0;
}