Minimum Inversion Number+线段树

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6529    Accepted Submission(s): 3979

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
 
#include<iostream>using namespace std;#define MAX 5000typedef struct infor{int left , right , value ;}infor ;infor tree[MAX*3];void create(int root, int left , int right ){tree[root].left = left ;tree[root].right = right ;tree[root].value = 0 ;if(right == left )return ;int mid = (left + right ) / 2 ;create(root*2 , left , mid);create(root*2 + 1 , mid + 1 , right );}void update(int root, int pos){tree[root].value += 1;if(tree[root].left == tree[root].right )return ;int mid = (tree[root].left + tree[root].right ) / 2 ;if(pos > mid )update(root*2 + 1, pos);else update(root*2 ,  pos);}int cal(int root, int left , int right ){if(tree[root].left == left && tree[root].right == right )return tree[root].value ;int mid = (tree[root].left + tree[root].right ) / 2;if(left > mid )return cal(root*2 + 1 , left , right );else if(right <= mid)return cal(root*2 , left , right );else return cal(root*2 + 1 , mid + 1, right ) + cal(root*2  , left , mid );}int main(){int n;while(cin >> n){int i ,a[MAX+1];create(1,0,n-1);int ans = 0 ;for( i = 0 ; i < n ; i ++){cin >> a[i] ;ans+=cal(1,a[i],n-1);update(1,a[i]);}int min = ans ;for(i = 0 ; i < n ; i ++){ans -=a[i];ans +=n-a[i] - 1;if(ans<min)min=ans;}cout<<min<<endl;}return 0;}

 
 
#include<iostream>using namespace std;#define MAX 5000int a[3*MAX],c[3*MAX]  , n ; int lowbit( int k ){return k&(-k) ;}void update(int sum , int k){while( k <= n ){c[k]+=sum ;k+=lowbit(k);}}int cal( int k){int sum = 0 ;while(k > 0 ){sum += c[k];k-=lowbit(k);}return sum ;}int main( void ){while( cin >> n ){memset(c,0,sizeof(c));int i;int ans=0;for( i = 1 ; i <= n ; i ++){cin >> a[i] ; a[i]++;ans+=cal(n)-cal(a[i]) ;update( 1 , a[i]);}int sum = ans ; for( i =1 ; i <= n ; i ++){ans+=n-a[i]-(a[i]-1) ;if(sum>ans)sum = ans ; }cout<< sum << endl ;}return 0 ;}