HDU 1394 线段树统计逆序对 解题报告

Minimum Inversion Number

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

【解题报告】

题意就是给出一串数，当依次在将第一个数变为最后一个数的过程中，要你求它的最小逆序数。

可以用线段数做，建的是一棵空树，然后每插入一个点之前，统计大于这个数的有多少个，直到所有的数都插入完成，就结果了逆序树的统计。

要得出答案主要是利用了一个结论，如果是0到n的排列，那么如果把第一个数放到最后，对于这个数列，逆序数是减少a[i],而增加n-1-a[i]的.当然，你也可以是统计小于这个数的有多少个，然后再用已经插入树中i个元素减去小于这个数的个数，得出的结果也是在一样的

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define N 5010

struct Node
{
int l,r;
int num;
}tree[N<<2];

void build(int i,int l,int r)
{
int mid=(l+r)>>1;
tree[i].l=l;tree[i].r=r;
tree[i].num=0;
if(l==r) return;
build(i<<1,l,mid);
build(i<<1|1,mid+1,r);
}
void updata(int i,int k)
{
if(tree[i].l==k&&tree[i].r==k)
{
tree[i].num=1;
return;
}
int mid=(tree[i].l+tree[i].r)>>1;
if(k<=mid) updata(i<<1,k);
else updata(i<<1|1,k);
tree[i].num=tree[i*2].num+tree[i*2+1].num;
}
int getsum(int i,int k,int n)
{
if(k<=tree[i].l&&tree[i].r<=n) return tree[i].num;
else
{
int mid=(tree[i].l+tree[i].r)>>1;
int sum1=0,sum2=0;
if(k<=mid) sum1=getsum(i<<1,k,n);
if(n>mid) sum2=getsum(i<<1|1,k,n);
return sum1+sum2;
}
}

int main()
{
int n;
while(scanf("%d",&n)!=EOF)
{
int a
;
build(1,0,n-1);
int i;
int ans=0;
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
ans+=getsum(1,a[i]+1,n-1);
updata(1,a[i]);
}
int minx=ans;
for(i=0;i<n;i++)
{
ans=ans+n-2*a[i]-1;
if(ans<minx) minx=ans;
}
printf("%d\n",minx);
}
return 0;
}