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

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

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