Ultra-QuickSort poj 2299--树状数组求逆序数

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 32150 Accepted: 11441

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

 

这个题比上一题更加明显，是裸的树状数组求逆序数，就是注意它的数据规模是999999999，要先进行离散化，再排序，求逆序数。

#include<iostream>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;const int maxn=500005;int c[maxn],n;//存树状数组,n表示点数int aa[maxn];//存离散化后的数组class node{    public:    int val,order;};node in[maxn];//存原始数据//树状数组的3个函数int lowbit(int x){    return x&(-x);}void update(int x,int val){    for(int i=x;i<=n;i+=lowbit(i))    {        c[i]+=val;    }}int getsum(int x){    int temp=0;    for(int i=x;i>=1;i-=lowbit(i))    {        temp+=c[i];    }    return temp;}bool cmp(node a,node b){    return a.val<b.val;}int main(){    while(scanf("%d",&n)==1&&n)    {        for(int i=1;i<=n;i++) {scanf("%d",&in[i].val);in[i].order=i;}        //离散化        sort(in+1,in+n+1,cmp);        for(int i=1;i<=n;i++) aa[in[i].order]=i;        //用树状数组求逆序数        memset(c,0,sizeof(c));        long long ans=0;        for(int i=1;i<=n;i++)        {            update(aa[i],1);            ans+=i-getsum(aa[i]);        }        cout<<ans<<endl;    }    return 0;}