HDU 1394 Minimum Inversion Number (线段树，单点更新)

C - Minimum Inversion Number
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Submit Status Practice HDU 1394

Appoint description:
System Crawler (2015-08-17)

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

题目大意：
　　这道题是说，给你一个长度为n的排列（由0到n-1组成），每次可以把第一个元素放到最后面，那么就会产生逆序数，一个排列的逆序数等于的是这个排列中的所有数字的逆序数的和。
那么，这道题，可以用线段树来解决，我们首先建立一个空树，也就是说，这个树中的每个节点的sum==0，表示全0，然后，我们每次插入一个数字，就在这个数字的对应位置放一个1，
也就是按照权值来建立这棵线段树，线段树最下面一层的叶子节点所维护的权值就是0~n-1。每次插入一个数字后，只要统计有多少个数字比这刚刚插入的这个数字大就可以了。
然后我们知道，对于一个仅仅由0~n-1组成的排列，把第一个数字放到最后一位，所产生的贡献是，减少了a[i],增加了n-1-a[i]个。那么每次都算一遍贡献，直到找到那个最小的贡献就可以了。

代码：

# include<cstdio>
# include<iostream>

using namespace std;

# define MAX 5004
# define inf 99999999
# define lid id<<1
# define rid id<<1|1

struct Segtree
{
int l,r;
int sum;
}tree[MAX*4];
int a[MAX];

void push_up( int id )
{
tree[id].sum = tree[lid].sum+tree[rid].sum;
}

void build ( int id,int l,int r )
{
tree[id].l = l; tree[id].r = r;
tree[id].sum = 0;
if ( l==r )
{
return;
}
int mid = ( tree[id].l+tree[id].r )/2;
build ( lid,l,mid);
build ( rid,mid+1,r);
push_up(id);
}

void update( int id,int x,int val )
{
if ( tree[id].l==tree[id].r )
{
tree[id].sum = 1;
return;
}
int mid = ( tree[id].l+tree[id].r )/2;
if ( x<=mid )
update(lid,x,val);
else
update(rid,x,val);
push_up(id);
}

int query( int id,int l,int r )
{
if ( tree[id].l==l&&tree[id].r==r )
{
return tree[id].sum;
}
int mid = ( tree[id].l+tree[id].r )/2;
if ( r <= mid )
return query(lid,l,r);
else if ( l > mid )
return query(rid,l,r);
else
{
return query(lid,l,mid)+query(rid,mid+1,r);
}
}

int main(void)
{
int n;
while ( scanf("%d",&n)!=EOF )
{
for ( int i = 0;i < n;i++ )
scanf("%d",&a[i]);
build(1,0,n-1);
int sum = 0;
for ( int i = 0;i < n;i++ )
{
if( a[i]!=n-1 )
{
sum+= query(1,a[i]+1,n-1);
update(1,a[i],1);
}
else
{
sum+=query(1,a[i],n-1);
update(1,a[i],1);
}
}
int ans = inf;
ans = min(ans,sum);
for ( int i = 0;i < n;i++ )
{
sum-=a[i];
sum+=n-1-a[i];
ans = min(ans,sum);
}
printf("%d\n",ans);
}

return 0;
}