[HDU POJ] 逆序数

HDU 1394

Minimum Inversion Number

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 11961    Accepted Submission(s): 7310

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

 

逆序数、改段求点

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
#define INF 0x3f3f3f3f
#define N 5010

int n;
int a[N];
int c[N];

int lowbit(int x)
{
    return x&(-x);
}
void update(int pos,int val)
{
    while(pos>0)
    {
        c[pos]+=val;
        pos-=lowbit(pos);
    }
}
int query(int pos)
{
    int res=0;
    while(pos<=n)
    {
        res+=c[pos];
        pos+=lowbit(pos);
    }
    return res;
}
int main()
{
    int res;
    while(scanf("%d",&n)!=EOF)
    {
        res=0;
        memset(c,0,sizeof(c));
        for(int i=1;i<=n;i++)
        {
            scanf("%d",&a[i]);
            a[i]++;
        }
        for(int i=1;i<=n;i++)
        {
            update(a[i]-1,1);
            res+=query(a[i]);
        }
        //printf("逆序数：%d",res);
        int Min=INF;
        for(int i=1;i<=n;i++)
        {
            res+=n-2*a[i]+1;
            Min=min(res,Min);
        }
        printf("%d
",Min);
    }
    return 0;
}

POJ 2299

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 43824		Accepted: 15983

Description

In this problem, you have to analyze a particular sorting algorithm. The algorithm processes a sequence of n distinct integers by swapping two adjacent sequence elements until the sequence is sorted in ascending order. For the input sequence 
9 1 0 5 4 ,
Ultra-QuickSort produces the output 
0 1 4 5 9 .
Your task is to determine how many swap operations Ultra-QuickSort needs to perform in order to sort a given input sequence.

Input

The input contains several test cases. Every test case begins with a line that contains a single integer n < 500,000 -- the length of the input sequence. Each of the the following n lines contains a single integer 0 ≤ a[i] ≤ 999,999,999, the i-th input sequence element. Input is terminated by a sequence of length n = 0. This sequence must not be processed.

Output

For every input sequence, your program prints a single line containing an integer number op, the minimum number of swap operations necessary to sort the given input sequence.

Sample Input

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0

Source

Waterloo local 2005.02.05

 

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cmath>
using namespace std;
#define ll long long
#define INF 0x3f3f3f3f
#define N 500000

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

int lowbit(int x)
{
    return x&(-x);
}
void update(int pos,int val)
{
    while(pos>0)
    {
        c[pos]+=val;
        pos-=lowbit(pos);
    }
}
int query(int pos)
{
    int res=0;
    while(pos<=N)
    {
        res+=c[pos];
        pos+=lowbit(pos);
    }
    return res;
}
int main()
{
    while(scanf("%d",&n),n)
    {
        ll res=0;
        memset(c,0,sizeof(c));
        for(int i=0;i<n;i++)
        {
            scanf("%d",&a[i]);
            b[i]=a[i];
        }
        sort(b,b+n);
        for(int i=0;i<n;i++)
        {
            int pos=lower_bound(b,b+n,a[i])-b;
            pos++;
            update(pos-1,1);
            res+=query(pos);
        }
        printf("%lld
",res);
    }
    return 0;
}