HDU 1394 Minimum Inversion Number （樹狀數組）

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 18379    Accepted Submission(s): 11158

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
        

Sample Output

16
        

Author CHEN, Gaoli Source ZOJ Monthly, January 2003 Recommend Ignatius.L   |   We have carefully selected several similar problems for you:  1698 1540 1542 1255 2795  題解： 問你不斷地将開頭的數放到結尾。求其中最小的逆序數。 樹狀數組求出最初的逆序數。複雜度O（nlogn），然後O(n)查詢其他解。 AC代碼：

#pragma comment(linker, "/STACK:102400000,102400000")
//#include<bits/stdc++.h>
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
#include <cstring>
#include <vector>
#include <map>
#include <cmath>
#include <queue>
#include <set>
#include <bitset>
#include <iomanip>
#include <list>
#include <stack>
#include <utility> 
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;
const double eps = 1e-8;  
const int INF = 1e9+7; 
const ll inf =(1LL<<62) ;
const int MOD = 1e9 + 7;  
const ll mod = (1LL<<32);
const int N =1e6+6; 
const int M=100010;
const int maxn=5010;
#define mst(a) memset(a, 0, sizeof(a))
#define M_P(x,y) make_pair(x,y)
#define in freopen("in.txt","r",stdin) 
#define rep(i,j,k) for (int i = j; i <= k; i++)  
#define per(i,j,k) for (int i = j; i >= k; i--)  
#define lson  l , mid , rt << 1    
#define rson  mid + 1 , r , rt << 1 | 1  
const int lowbit(int x) { return x&-x; }
//const int lowbit(int x) { return ((x)&((x)^((x)-1))); } 
int read(){ int v = 0, f = 1;char c =getchar();
while( c < 48 || 57 < c ){if(c=='-') f = -1;c = getchar();}
while(48 <= c && c <= 57) v = v*10+c-48, c = getchar();
return v*f;}
int a[maxn];
int n,m;
int c[maxn];
void add(int i,int val)
{
	while(i<=n)
	{
		c[i]+=val;
		i+=lowbit(i);
	}
}
int sum(int i)
{
	int s = 0;
	while(i>0)
	{
		s+=c[i];
		i-=lowbit(i);
	}
	return s;
}
int main()
{
	
	while(~scanf("%d",&n))
	{
		int ans = 0;
		memset(c,0,sizeof(c)); 
		for(int i=1;i<=n;i++)
		{
			scanf("%d",&a[i]);
			a[i]++;
			ans+=sum(n)-sum(a[i]);
			add(a[i],1);
		}
		int MIN = ans;
		for(int i=1;i<=n;i++)
		{
			ans-=a[i];//減少的逆序數 
			ans+=n-a[i]+1; //增加的逆序數 
			if(ans<MIN) MIN=ans;
		}
		cout<<MIN<<endl;
	}
    return 0;
}