Minimum Inversion Number （单点更新 线段树 ）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 19112 Accepted Submission(s): 11525



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

总共有N个数，如何判断第i+1个数到最后一个数之间有多少个数小于第i个数呢？不妨假设有一个区间 [1,N]，只需要判断区间[i+1,N]之间有多少个数小于第i个数。如果我们把总区间初始化为0，然后把第i个数之前出现过的数都在相应的区间把它的值定为1，那么问题就转换成了[i+1,N]值的总和。再仔细想一下，区间[1,i]的值+区间[i+1,N]的值=区间[1,N]的值(i已经标记为1)，所以区间[i+1,N]值的总和等于N-[1,i]的值！因为总共有N个数，不是比它小就是比它(大或等于)。

现在问题已经转化成了区间问题，枚举每个数，然后查询这个数前面的区间值的总和，i-[1,i]既为逆序数。

线段树预处理时间复杂度O(NlogN)，N次查询和N次插入的时间复杂度都为O(NlogN)，总的时间复杂度O(3*NlogN)

/*num数组 是把记录 数是否存在 存在即为1。
总共有N个数，如何判断第i+1个数到最后一个
数之间有多少个数小于第i个数呢？不妨假设
有一个区间 [1,N]，只需要判断区间[i+1,N]之
间有多少个数小于第i个数。如果我们把总区间初
始化为0，然后把第i个数之前出现过的数都在相应
的区间把它的值定为1，那么问题就转换成了[i+1,N]
值的总和
*/
#include <stdio.h>
#define lson l, m, rt<<1
#define rson m+1, r, rt<<1|1
#define size 5005
#define min(a, b) a > b ? b : a
int num[size << 2], x[size];
void pushup(int rt)
{
num[rt] = num[rt << 1] + num[rt << 1 | 1];
}
void build( int l, int r, int rt)
{
num[rt] = 0;
if( l == r) return;
int m = (l + r) >> 1;
build(lson);
build(rson);
}

int qurey( int L, int R, int l, int r, int rt)
{
if( L <= l && r <= R)
return num[rt];
int m = (l + r) >> 1;
int ans = 0;
if(L <= m) ans+=qurey(L, R, lson);
if(R > m) ans+=qurey(L, R, rson);
return ans;
}
void updata( int p, int l, int r, int rt)
{
if( l == r)
{
num[rt]++;return;
}
int m = ( l + r) >> 1;
if( p <= m) updata(p, lson);
else updata(p, rson);
pushup(rt);
}
int main()
{
int n;
while(~scanf("%d", &n))
{
int sum = 0;
build(0, n - 1, 1);
for( int i= 1; i <= n; i++)
{
scanf("%d", &x[i]);
sum += qurey(x[i], n - 1, 0, n - 1, 1);
updata(x[i], 0, n - 1, 1);

}
int tmp = sum;
for( int i = 1; i <= n; i++)
{
sum += n - x[i] - x[i] - 1;
tmp = min(tmp, sum);
}
printf("%d\n",tmp);
}
return 0;
}