1394 ( Minimum Inversion Number ) (线段树)-优快云博客

本文链接：https://blog.youkuaiyun.com/Burning_newbie/article/details/51025731

本文介绍了一种通过线段树解决最小逆序数问题的方法。该问题要求在一系列由初始排列衍生的序列中找到逆序数最少的那个。文章提供了一个完整的C++实现示例，并解释了如何使用线段树来高效地查询和更新逆序数。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

http://acm.hdu.edu.cn/showproblem.php?pid=1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16259 Accepted Submission(s): 9891

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

线段树水题，输入一个x，查询【x，n-1】区间有多少数，然后update【x】，如果先更新，那么查询的值就是【x+1，n-1】了，可能会出错

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <map>
#include <set>
#include <stack>
#include <string>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
using namespace std;
const int SIZE=5e3+10;
const int maxn=500000;
const int Mod=1e8;
int sum[SIZE<<2];
int a[SIZE];
void pushup(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void build(int l,int r,int rt){
    memset(sum,0,sizeof(sum));
}
void update(int x,int l,int r,int rt){
    if(l==r){
        sum[rt]=1;
        return;
    }
    int m=(l+r)>>1;
    if(x<=m)update(x,lson);
    else update(x,rson);
    pushup(rt);
}
int query(int x,int l,int r,int rt){
    if(x<=l){
        return sum[rt];
    }
    int m=(l+r)>>1;
    int cnt=0;
    if(x<m)cnt+=query(x,lson)+query(m+1,rson);
    else cnt+=query(x,rson);
    return cnt;
}
int main()
{
    int n;
    while(scanf("%d",&n)!=EOF){
        int sum=0;
        build(0,n-1,1);
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            sum+=query(a[i],0,n-1,1);
            update(a[i],0,n-1,1);
        }
        int ans=sum;
        for(int i=0;i<n;i++){
            sum+=n-1-2*a[i];
            ans=min(ans,sum);
        }
        printf("%d\n",ans);
    }
    return 0;
}