Minimum Inversion Number（线段树求逆序数）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 15517    Accepted Submission(s): 9467

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

10 1 3 6 9 0 8 5 7 4 2

mergesort

 

Sample Output

16

 

Author

CHEN, Gaoli

题解：线段树；也可以用归并排序，也可以用树状数组；注意每次把第一个放在最后这个条件；归并：http://www.cnblogs.com/handsomecui/p/4814442.html

mergesort

代码：

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<algorithm>
#include<vector>
#include<map>
using namespace std;
const int MAXN=5010;
#define lson root<<1,l,mid
#define rson root<<1|1,mid+1,r
#define pushup tree[root]=tree[root<<1]+tree[root<<1|1]
#define mem(x,y) memset(x,y,sizeof(x))
int tree[MAXN<<2];
int a[MAXN];
int ans; 
void update(int v,int root,int l,int r){
	int mid=(l+r)>>1;
	if(l==v&&r==v){
		tree[root]++;
		return;
	}
	if(mid>=v)update(v,lson);
	if(mid<v)update(v,rson);
	pushup;
}
void query(int L,int R,int root,int l,int r){
	int mid=(l+r)>>1;
	if(l>=L&&r<=R){
		ans+=tree[root];
		return;
	}
	if(mid>=L)query(L,R,lson);
	if(mid<R)query(L,R,rson);
}
int main(){
	int N;
	while(~scanf("%d",&N)){
		mem(tree,0);
		ans=0;
		int x;
		for(int i=0;i<N;i++){
			scanf("%d",&x);a[i]=x;
			query(x+1,N-1,1,0,N-1);
		//	printf("%d
",ans);
			update(x,1,0,N-1);
		}
		int cnt=ans;
		for(int i=0;i<N;i++){
			cnt=cnt+N-1-a[i]-a[i];
			ans=min(ans,cnt);
		}
		printf("%d
",ans);
	}
	return 0;
}