[原创]HDU 1394 Minimum Inversion Number [线段树->单点更新]【数据结构】

2016-11-13 16:27:43 Tabris_ 阅读数：237

博客爬取于2020-06-14 22:42:38
以下为正文

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https://blog.csdn.net/qq_33184171/article/details/53149713

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

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Minimum Inversion Number
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 18798 Accepted Submission(s): 11367

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

Author
CHEN, Gaoli

Source
ZOJ Monthly, January 2003

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题目大意：
就是有一个序列，这个序列可以循环，变成下列这么些个序列。
a1, a2, …, an-1, an (m = 0 - the initial seqence)
a2, a3, …, an, a1 (m = 1)
a3, a4, …, an, a1, a2 (m = 2)
…
an, a1, a2, …, an-1 (m = n-1)
然后对于每种序列计算逆序对数
然后输出逆序对数最小的值

解题思路:
首先对于最初的序列求解逆序对数，只需要线段树or树状数组就行了
这里采取的是线段树维护

在每次把数据挂到树上之前可以区间查询的方式计算这个值的逆序数然后O( $n$ )遍历一遍就能知道整个序列的逆序数了
总复杂度是O( $nlogn$ )

然后他要求的序列一共有n个那样的话不能O( $n^2logn$ ) 这样复杂度实在太高了

然后想最后发现了有一个规律；
因为序列中的数就是0~n-1且新的序列是从旧的序列移过来的。
那么逆序数就会相应减少 $a_i$ 个同时就增加了 $n-a_i-1$ 个
这样的话就能够O( $1$ )处理了

附本题代码
------------------------------------.

# include <bits/stdc++.h>

# define abs(x)          (((x)>0)?(x):-(x))
# define lalal           puts("*********")
# define Rep(a,b,c)      for(int a=(b);a<=(c);a++)
# define Req(a,b,c)      for(int a=(b);a>=(c);a--)
# define Rop(a,b,c)      for(int a=(b);a<(c);a++)
# define s1(a)           scanf("%d",&a)
typedef long long int LL;
using namespace std;
const int inf = 0x3f3f3f3f;
const int MOD = 9901;
/**************************************/
const int N = 10000+5;
struct node
{
    int l,r;
    int val;
    int md()
    {
        return (l+r)>>1;
    }
}tree[N<<2];
int a[N],ans;
# define ll  (rt<<1)
# define rr  (rt<<1|1)
# define mid (tree[rt].md())
void pushup(int rt)
{
    tree[rt].val=tree[ll].val+tree[rr].val;
}
void build(int rt,int l,int r)
{
    tree[rt].l=l,tree[rt].r=r;
    if(l==r)
    {
        tree[rt].val=0;
        return ;
    }
    build(ll,l,mid);
    build(rr,mid+1,r);
    pushup(rt);

    return;
}

void update(int rt,int pos,int val)
{
    if(tree[rt].l==tree[rt].r)
    {
        tree[rt].val+=val;
        return;
    }

    if(pos<=mid)update(ll,pos,val);
    else      update(rr,pos,val);
    pushup(rt);
    return;
}

void query(int rt,int L,int R)
{
    if(L<=tree[rt].l&&tree[rt].r<=R)
    {
        ans += tree[rt].val;
        return;
    }
    if(R<=mid)
        query(ll,L,R);
    else if(L>mid)
        query(rr,L,R);
    else
    {
        query(ll,L,R);
        query(rr,L,R);
    }
    return;
}

int main()
{
    int n;
    while(~s1(n))
    {
        build(1,0,n);

        int sum = 0;
        Rep(i,1,n)
        {
            s1(a[i]);
            ans = 0;
            query(1,a[i]+1,n);
            sum+=ans;
            update(1,a[i],1);
        }

        int mi = sum;
        Rep(i,1,n)
        {
            sum+=n-a[i]-a[i]-1;
            if(sum<mi)
                mi=sum;
        }
        printf("%d\n",mi);
    }
    return 0;
}