Minimum Inversion Number 线段树+单点修改

最新推荐文章于 2019-02-25 23:17:34 发布

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最新推荐文章于 2019-02-25 23:17:34 发布

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分类专栏：线段树

本文链接：https://blog.youkuaiyun.com/u012870383/article/details/38038269

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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变化而来的序列中找到逆序数最少的那个序列。文章提供了一个具体的C++实现案例，并解释了每次移动元素如何影响逆序数的变化。

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Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 10117 Accepted Submission(s): 6203

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

解决方案：hdu上正确率60+%的题目还要看题解.......，看来线段树没学到家啊，好吧！其实这道题也可以暴力。本题是求逆序，可先求原始装态的逆序个数，然后对于每次移动，逆序个数会变成cnt+=N-a[i]-a[i]-1。即每次把第一个数移到最后，会增加N-a[i]-1，同时会减少a[i]对逆序。求初识逆序可用线段树。

code:

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstring>
#define MMAX  5005
using namespace std;
int sum[4*MMAX],N,_min,_ant;
int _sum[MMAX],y[MMAX],a[MMAX];
void build(int rt,int L,int R)
{
    if(L==R)
    {
        sum[rt]=0;
    }
    else
    {
        int M=(L+R)/2;
        build(rt*2,L,M);
        build(rt*2+1,M+1,R);
        sum[rt]=0;

    }

}
void query(int rt,int L,int R,int st,int en)
{

    if(st<=L&&en>=R)
    {
        _ant+=sum[rt];
    }
    else
    {
        int M=(L+R)/2;
       //cout<<rt<<endl;
        //getchar();
        if(st<=M) query(rt*2,L,M,st,en);
        if(en>M) query(rt*2+1,M+1,R,st,en);

    }


}
void update(int rt,int L,int R,int C)
{

    if(L==R)
    {
       sum[rt]++;
    }
    else
    {
        int M=(L+R)/2;
        if(C<=M)update(rt*2,L,M,C);
        else update(rt*2+1,M+1,R,C);
        sum[rt]=sum[rt*2]+sum[rt*2+1];
    }

}
int main()
{
    while(~scanf("%d",&N))
    {
        build(1,1,N);
        _min=0;
        for(int i=1;i<=N;i++)
            scanf("%d",&a[i]);
        for(int i=1; i<=N; i++)
        {
            _ant=0;
            query(1,1,N,a[i]+1,N);
            update(1,1,N,a[i]+1);
            _min+=_ant;
           // cout<<_ant<<endl;
        }
        int min_=_min;
        for(int i=1; i<=N; i++)
        {
            _min+=N-2*a[i]-1;

            min_=min(_min,min_);
        }
        printf("%d\n",min_);
    }
    return 0;
}