POJ 2299 Ultra-QuickSort

离散化+树状数组求逆序数

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 35024		Accepted: 12608

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 ≤ 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 <cstdio> #include <cstring> #include <algorithm> using  namespace std; struct node {      int v,ord; }s[ 550000]; bool cmp(node a,node b) {      return a.v<b.v; } int ar[ 550000],tree[ 550000],n; int lowbit( int x) {      return x&-x; } void update( int x, int v) {      while(x<=n)     {         tree[x]+=v;         x+=lowbit(x);     } } int getsum( int x) {      int sum= 0;      while(x> 0)     {         sum+=tree[x];         x-=lowbit(x);     }      return sum; } int main() {      while(scanf( "%d",&n)&&n)     {          for( int i= 1;i<=n;i++)         {             scanf( "%d",&s .v);             s.ord=i;         }         sort(s+1,s+n+1,cmp);         for(int i=1;i<=n;i++)             ar[s.ord]=i;         memset(tree,0,sizeof(tree));         long long int ans=0;         for(int i=1;i<=n;i++)         {             update(ar,1);             ans+=i-getsum(ar);         }         printf("%I64d\n",ans);     }     return 0; }

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