POJ2299-Ultra-QuickSort

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 56022 Accepted: 20697

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

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05

题意：给出长度为n的序列，每次只能交换相邻的两个元素，问至少要交换几次才使得该序列为递增序列。

解题思路：一个乱序序列的 逆序数 = 在只允许相邻两个元素交换的条件下,得到有序序列的交换次数

#include <iostream>#include <cstdio>#include <cstring>#include <stack>#include <queue>#include <cmath>#include <algorithm>using namespace std;const int maxn=501000;struct node{    long long int x;    int y;}a[maxn];int n,b[maxn],t;long long int f[maxn];bool cmp(node a,node b){    return (a.x<b.x||(a.x==b.x && a.y<b.y));}int lowb(int x){    return x&(-x);}long long q(int x){    long long int tmp=0;    while (x>0)    {        tmp+=f[x];        x-=lowb(x);    }    return tmp;}void add(int x){    while(x<=n)    {        f[x]++;        x+=lowb(x);    }}int main(){    while(~scanf("%d",&n)&&n)    {        for (int i=1; i<=n; i++)        {            scanf("%I64d",&a[i].x);            a[i].y=i;        }        t=0;        sort(a+1,a+1+n,cmp);        a[0].x=-1;        for (int i=1;i<=n;i++)        {            if (a[i].x!=a[i-1].x) t++;            b[a[i].y]=t;        }        memset(f,0,sizeof(f));        long long int ans=0;        for (int i=1;i<=n;i++)        {            ans+=(q(n)-q(b[i]));            add(b[i]);        }        cout<<ans<<endl;    }    return 0;}