Minimum Inversion Number（线段树）

Minimum Inversion Number

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

Total Submission(s): 15151    Accepted Submission(s): 9250


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

 

题意：

       给你N个数，要求统计它的所有形式的逆序对的最小值。它的所有形式的意思是，不断将数组开头的第一个数放到数组的最后面。
这题求逆序数的用了我四五个晚上终于A了，一开始参考别人的代码，结果却一直运行不起来。到运行起来了，却答案错误。现在终于将它A了，很高兴。

AC代码：

#include<iostream>
#include<algorithm>
#include<cstring>
#include<cstdio>
using namespace std;
#define T 5005
int n, a[T],ans;
struct node
{
int l,r,mid;
int v;
}tree[T<<2];
void PushUp(int rt)
{
tree[rt].v = tree[rt << 1].v + tree[rt << 1 | 1].v;
}
void Build(int rt,int L,int R)
{
tree[rt].l = L;
tree[rt].r = R;
tree[rt].mid = (L + R) >> 1;
tree[rt].v = 0;
if (L == R)return;
Build(rt<<1,L,tree[rt].mid);
Build(rt<<1|1,tree[rt].mid+1,R);
}
void add(int rt,int pos)
{
if (tree[rt].l ==pos && tree[rt].r==pos){
tree[rt].v=1;
return;
}
if (pos <= tree[rt].mid){
add(rt<<1,pos);
}
else
{
add(rt<<1|1,pos);
}
PushUp(rt);
}
void query(int rt,int L,int R)
{
if (tree[rt].l>=L&&tree[rt].r<=R){
ans+=tree[rt].v; return;
}
if (R <= tree[rt].mid){
query(rt<<1,L,R);
}
else if (L > tree[rt].mid){
query(rt<<1|1,L,R);
}
else
{
query(rt<<1,L,tree[rt].mid);
query(rt<<1|1,tree[rt].mid+1,R);
}
}
int main()
{
/*freopen("input.txt","r",stdin);*/
int i;
while (~scanf("%d", &n))
{
ans = 0;
Build(1,1,n);
for (i = 0; i < n;++i){
scanf("%d",&a[i]);
a[i]++;
if (a[i]!=n)
query(1,a[i]+1,n);
add(1,a[i]);
}
int mi = ans;
for (i = 0; i < n;++i){
ans = ans - ((a[i]-1)<<1) + n - 1;
mi = min(mi,ans);
}
printf("%d\n",mi);
}
return 0;
}