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

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操作后找到给定序列的所有可能排列中逆序数最小的一个。文章详细解释了如何使用线段树来高效地计算每次操作后的逆序数。

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18913 Accepted Submission(s): 11420

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：
给出n个数，n次操作，每次都把第一个元素放到最后。
问形成的n个数组中的最小逆序数。

解题思路：
如果已知某一数组的逆序数，那么把第i个元素放到最后，新数组的逆序数为
sum + n - 1 - a[i]*2

那么现在是求对于每一个插入的数，已有的比他大的数的个数。
就用线段树去搜索，节点记录下区间里的数出现的总次数。

AC代码:

#include<bits/stdc++.h>
#define lson l,mid,rt<<1
#define rson mid+1,r,rt<<1|1
using namespace std;
const int maxn = 5e3;
int stree[maxn<<1];
int a[maxn];
void PushUp(int rt) {stree[rt] = stree[rt<<1]+stree[rt<<1|1];}
void build(int l,int r,int rt)
{
    stree[rt] = 0;
    if(l == r)  return ;
    int mid = (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 stree[rt];
    int mid = (l+r)>>1;
    int res = 0;
    if(L <= mid)    res += query(L,R,lson);
    if(R > mid)     res += query(L,R,rson);
    return res;
}
void Update(int l,int r,int rt,int c)
{
    if(l == r)
    {
        stree[rt]++;
        return ;
    }
    int mid = (l+r)>>1;
    if(c <= mid)    Update(lson,c);
    else            Update(rson,c);
    PushUp(rt);
}
int main()
{
    int n;
    while(~scanf("%d",&n))
    {
        build(1,n,1);
        int sum = 0;
        for(int i =  0;i < n;i++)
        {
            scanf("%d",&a[i]);
            sum += query(a[i]+1,n-1,0,n-1,1);
            Update(0,n-1,1,a[i]);
        }
        int res = INT_MAX;
        for(int i = 0;i < n;i++)    sum += n-1-a[i]*2, res = min(res,sum);
        printf("%d\n",res);
    }
    return 0;
}