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.
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.
InputThe 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.
OutputFor 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
这题先要补充一个逆序数的概念
在一个排列中，如果一对数的前后位置与大小顺序相反，即前面的数大于后面的数，
那么它们就称为一个逆序。一个排列中逆序的总数就称为这个排列的逆序数。
一个排列中所有逆序总数叫做这个排列的逆序数。也就是说，对于n个不同的元素，
先规定各元素之间有一个标准次序（例如n个 不同的自然数，可规定从小到大为标准次序），
于是在这n个元素的任一排列中，当某两个元素的先后次序与标准次序不同时，
就说有1个逆序。一个排列中所有逆序总数叫做这个排列的逆序数。
(以上文字引用自百度百科）
由于数据较大求逆序数就必须用复杂度较少的方法求，强行暴力求逆序数表示应该没人会选择这样做吧。
线段树就是一个非常好的数据结构用于求逆序数。
题意：一个有N个数的序列中，每次把第一个放在序列的最后一个形成新的序列，
求这些序列中最小的逆序数是多少？
求出原始序列的逆序数S，将a[i]移到最后一位时，新的序列的逆序数为
原来序列的逆序值S+比a[i]大的数的个数（n-1-a[i])-a[i]（比a[i]小的个数）
如果这题本菜鸟有什么地方理解不正确，请读者在下方给我留言，纠正我的错误。
1 #include<iostream> 2 #include<stdio.h> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 #define maxn 5010 7 int sum[maxn*4+10],a[maxn]; 8 void pushup(int rt) 9 { 10 sum[rt]=sum[rt<<1]+sum[rt<<1|1]; 11 } 12 void build(int l,int r,int rt) 13 { 14 sum[rt]=0; 15 if (l==r) return ; 16 int m=(l+r)>>1; 17 build(l,m,rt<<1); 18 build(m+1,r,rt<<1|1); 19 } 20 void updata(int x,int l,int r,int rt ) 21 { 22 if (l==r) { 23 sum[rt]++; 24 return ; 25 } 26 int m=(l+r)>>1; 27 if (x<=m) updata(x,l,m,rt<<1); 28 else updata(x,m+1,r,rt<<1|1); 29 pushup(rt); 30 } 31 int query(int x,int y,int l,int r,int rt) 32 { 33 if (x<=l && r<=y) return sum[rt]; 34 int m=(l+r)>>1; 35 int ans=0; 36 if (x<=m) ans+=query(x,y,l,m,rt<<1); 37 if (y>m) ans+=query(x,y,m+1,r,rt<<1|1); 38 return ans; 39 } 40 int main() 41 { 42 int n; 43 while(scanf("%d",&n)!=EOF){ 44 build(0,n-1,1); 45 int s=0; 46 for (int i=0 ;i<n ;i++){ 47 scanf("%d",&a[i]); 48 s+=query(a[i],n-1,0,n-1,1); 49 updata(a[i],0,n-1,1); 50 } 51 int ans=s; 52 for (int i=0 ;i<n ;i++ ){ 53 s+=(n-a[i]-1)-a[i]; 54 ans=min(ans,s); 55 } 56 printf("%d ",ans); 57 } 58 return 0; 59 }