Minimum Inversion Number
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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
101 3 6 9 0 8 5 7 4 2
Sample Output
16
Author
#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>#define MAX 5050using namespace std;struct Node{ int l,r,sum;}node[MAX<<2];void build_tree(int l,int r,int root){ node[root].l=l; node[root].r=r; if(l==r) { node[root].sum=0; return ; } int mid=(l+r)>>1; build_tree(l,mid,root<<1); build_tree(mid+1,r,(root<<1)|1); node[root].sum=0;}void add(int i,int t,int val){ node[i].sum+=val; if(node[i].l==node[i].r){ return ; } int mid=(node[i].l+node[i].r)>>1; if(t<=mid)add(i<<1,t,val); else add((i<<1)|1,t,val);}int sum(int l,int r,int root){ if(node[root].l==l&&node[root].r==r) return node[root].sum; int mid=(node[root].r+node[root].l)>>1; if(r<=mid)return sum(l,r,root<<1); else if(l>mid)return sum(l,r,(root<<1)|1); else { return sum(l,mid,root<<1)+sum(mid+1,r,(root<<1)|1); }}int a[MAX];int main(){ int n; while(~scanf("%d",&n)) { build_tree(1,n,1); for(int i=1;i<=n;i++) scanf("%d",&a[i]); int ans=0; for(int i=1;i<=n;i++) { ans+=sum(a[i]+1,n,1); add(1,a[i]+1,1); } int Min=ans; for(int i=1;i<=n;i++){ ans-=a[i]; ans+=n-a[i]-1; if(ans<Min)Min=ans; } printf("%d\n",Min); } return 0;}
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