HDU 1394 树状数组求逆序对

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13036    Accepted Submission(s): 7968

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

 

题目意思：

给一个n，然后一个长度为n的数组，每个元素小于n且不重复，每次操作把首位元素放到末尾，求n次操作中最小的逆序对数目。

 

思路：

先求一下原始数组的逆序对数目，每次把首位放末尾的时候，num（逆序对数目）加上大于a[i]的，然后减去小于a[i]的。

 

代码：

#include <cstdio>
#include <cstring>
#include <algorithm>
#include <iostream>
#include <vector>
#include <queue>
#include <cmath>
#include <set>
using namespace std;

#define N 5005
#define ll root<<1
#define rr root<<1|1
#define mid (a[root].l+a[root].r)/2


int max(int x,int y){return x>y?x:y;}
int min(int x,int y){return x<y?x:y;}
int abs(int x,int y){return x<0?-x:x;}

int n;
int b[N];
int a[N];

int lowbit(int x){
    return x&(-x);
}

void solve(int val,int x){
    while(x<=n){
        a[x]+=val;
        x+=lowbit(x);
    }
}

int sum(int x){
    int ans=0;
    while(x>0){
        ans+=a[x];
        x-=lowbit(x);
    }
    return ans;
}

main()
{
    int t, i, j, k;
    while(scanf("%d",&n)==1){
        memset(a,0,sizeof(a));
        int ans=0;
        for(i=1;i<=n;i++) {
            scanf("%d",&b[i]);
            b[i]++;
            ans+=sum(n)-sum(b[i]);
            solve(1,b[i]);
        }
        int num=ans;
        for(i=1;i<=n;i++){
            num+=n-b[i]-b[i]+1;//sum(n)-sum(b[i])-sum(b[i]-1);
            ans=min(ans,num);
        }
        printf("%d
",ans);
    }
}