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
#include <stdio.h>
#include <algorithm> using namespace std; int a[];
struct Node{
int l,r,num;
}tree[]; void Build(int n,int x,int y){
tree[n].l = x;
tree[n].r = y;
tree[n].num = ;
if(x == y){
return;
}
int mid = (x + y) / ;
Build(*n,x,mid);
Build(*n+,mid+,y);
} void Modify(int n,int x){
int l = tree[n].l;
int r = tree[n].r;
int mid = (l + r) / ;
if(x == l && x == r){
tree[n].num = ;
return;
}
if(x <= mid) Modify(*n,x);
else Modify(*n+,x);
tree[n].num = tree[*n].num + tree[*n+].num;
} int Query(int n,int x,int y){
int l = tree[n].l;
int r = tree[n].r;
int mid = (l + r) / ;
int ans = ;;
if(x == l && y == r)
return tree[n].num;
if(x <= mid) ans += Query(*n,x,min(mid,y));
if(y > mid) ans += Query(*n+,max(mid+,x),y);
return ans;
}
int main(){
//freopen ("a.txt" , "r" , stdin ) ;
int n,sum,ans;
int i,j; while(scanf("%d",&n) != EOF){
sum = ;
Build(,,n);
for(i = ;i <= n;i++){
scanf("%d",&a[i]);
Modify(,a[i]);
sum += Query(,a[i]+,n);
}
ans = sum;
for(i = ;i < n;i++){
sum = sum + (n - - a[i]) - a[i];
if(sum < ans)
ans = sum;
}
printf("%d\n",ans);
}
}