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.
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
题目大意：求逆序对数目，然后将第一个数字置于队尾在求一遍，直到每个数字都到过队尾的最小逆序对数.
思路：用线段树维护，每次插入统计在其右边的逆序对数和，再用公式推导出每次变换的逆序对数。
attention:注意“0”的存在，被坑了好几次。
1 #include <iostream> 2 #include <cstdio> 3 #include <cmath> 4 #include <cstring> 5 #define MAXN 100000 6 using namespace std; 7 int segtree[MAXN*4],a[MAXN*4]; 8 int n,sum; 9 void adddata(int now) 10 { 11 segtree[now]=segtree[(now<<1)]+segtree[(now<<1)+1]; 12 } 13 void buildtree(int now,int l,int r) 14 { 15 segtree[now]=0; 16 if (l==r) return; 17 int mid=(l+r)>>1; 18 buildtree((now<<1),l,mid); 19 buildtree((now<<1)+1,mid+1,r); 20 adddata(now); 21 } 22 int query(int now,int l,int r,int begin,int end) 23 { 24 if (begin<=l && end>=r) return segtree[now]; 25 int mid=(l+r)>>1,ans=0; 26 if (begin<=mid) ans+=query((now<<1),l,mid,begin,end); 27 if (end>mid) ans+=query((now<<1)+1,mid+1,r,begin,end); 28 return ans; 29 } 30 void pointchange(int now,int l,int r,int x,int v) 31 { 32 if (l==r) {segtree[now]+=v; return;} 33 int mid=(l+r)>>1; 34 if (x<=mid) pointchange((now<<1),l,mid,x,v); 35 else pointchange((now<<1)+1,mid+1,r,x,v); 36 adddata(now); 37 } 38 int main() 39 { 40 int i; 41 int minn; 42 while (~scanf("%d",&n)) 43 { 44 buildtree(1,0,n-1); 45 sum=0; 46 for (i=0;i<n;i++) 47 { 48 scanf("%d",&a[i]); 49 sum+=query(1,0,n-1,a[i],n-1 ); 50 pointchange(1,0,n-1,a[i],1); 51 } 52 minn=sum; 53 for (i=0;i<n;i++) 54 { 55 sum+=n-2*a[i]-1; 56 minn=min(minn,sum); 57 } 58 printf("%d ",minn); 59 } 60 return 0; 61 }