Minimum Inversion Number
Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u
Submit Status
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
1 #include <stdio.h>
2 #include <algorithm>
3
4 using namespace std;
5
6 int a[5005];
7 struct Node{
8 int l,r,num;
9 }tree[50000];
10
11 void Build(int n,int x,int y){
12 tree[n].l = x;
13 tree[n].r = y;
14 tree[n].num = 0;
15 if(x == y){
16 return;
17 }
18 int mid = (x + y) / 2;
19 Build(2*n,x,mid);
20 Build(2*n+1,mid+1,y);
21 }
22
23 void Modify(int n,int x){
24 int l = tree[n].l;
25 int r = tree[n].r;
26 int mid = (l + r) / 2;
27 if(x == l && x == r){
28 tree[n].num = 1;
29 return;
30 }
31 if(x <= mid) Modify(2*n,x);
32 else Modify(2*n+1,x);
33 tree[n].num = tree[2*n].num + tree[2*n+1].num;
34 }
35
36 int Query(int n,int x,int y){
37 int l = tree[n].l;
38 int r = tree[n].r;
39 int mid = (l + r) / 2;
40 int ans = 0;;
41 if(x == l && y == r)
42 return tree[n].num;
43 if(x <= mid) ans += Query(2*n,x,min(mid,y));
44 if(y > mid) ans += Query(2*n+1,max(mid+1,x),y);
45 return ans;
46 }
47 int main(){
48 //freopen ("a.txt" , "r" , stdin ) ;
49 int n,sum,ans;
50 int i,j;
51
52 while(scanf("%d",&n) != EOF){
53 sum = 0;
54 Build(1,0,n);
55 for(i = 1;i <= n;i++){
56 scanf("%d",&a[i]);
57 Modify(1,a[i]);
58 sum += Query(1,a[i]+1,n);
59 }
60 ans = sum;
61 for(i = 1;i < n;i++){
62 sum = sum + (n - 1 - a[i]) - a[i];
63 if(sum < ans)
64 ans = sum;
65 }
66 printf("%d\n",ans);
67 }
68 }