Minimum Inversion Number（线段树）_minimum inversion number线段树-优快云博客

本文介绍了一种利用线段树求解给定序列及其循环移位后所有序列中最小逆序对数量的方法。通过计算初始序列的逆序对数量，并结合移位操作对逆序对数量的影响规律，高效地找出所有可能序列中逆序对数量的最小值。

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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

解题思路：

用线段树求逆序对。只需要求出第一组逆序对就能推出移项之后的逆序对。假设一组序列有n个元素，元素包含0 ~ n - 1，未移项时逆序对有m组，第一个元素为a，那么当第一个元素移到最后时，比第一个元素小的元素都失去了这组逆序对，所以少了c组逆序对，而比第一个元素大的元素都多了一组逆序对，所以多了 n - 1 - c。因此移项一次项后的逆序对为m - c + (n - 1 - c)组。通过这个公式不断递推即可求得每次移项后的逆序对个数，从而求出最小的个数。

AC代码：

#include <iostream>
#include <cstdio>
using namespace std;
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
const int maxn = 5005;
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 m = (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 m = (l + r) >> 1;
    if(p <= m)
        update(p, lson);
    else
        update(p, rson);
    PushUp(rt);
}
int query(int L, int R, int l, int r, int rt)
{
    if(L <= l && r <= R)
        return sum[rt];
    int m  = (l + r) >> 1;
    int ret = 0;
    if(L <= m)
        ret += query(L, R, lson);
    if(m < R)
        ret += query(L, R, rson);
    return ret;
}
int main()
{
    int n, a[maxn], sum, ans;
    while(scanf("%d", &n) != EOF)
    {
        sum = 0;
        build(0, n - 1, 1);
        for(int i = 0; i < n; i++)
        {
            scanf("%d", &a[i]);
            sum += query(a[i] + 1, n - 1, 0, n - 1, 1);
            update(a[i], 0, n - 1, 1);
        }
        ans = sum;
        for(int i = 0; i < n - 1; i++)
        {
            sum += n - 1 - 2 * a[i];
            ans = min(ans, sum);
        }
        printf("%d\n", ans);
    }
    return 0;
}