poj 2299 Ultra-QuickSort (树状数组解法)

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 32746 Accepted: 11691

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

ps：刚学了下树状数组  所以这题用树状数组做的  不过这题正规方法应该用归并排序做的 

    各位亲们 想了解 树状数组的可以看一下 不想的就直接pass哈

#include<cstdio>#include <cstring>#include <algorithm>#define maxn 500005using namespace std;int n;__int64 ans;int a[maxn];struct Tnode{    int num;    int val;}node[maxn];int lowbit(int v){    return v&(-v);}void update(int v){    for(int i=v; i<=maxn-5; i+=lowbit(i))    {        a[i]++;    }}int getsum(int v){    int sum=0;    for(int i=v; i>0; i-=lowbit(i))    {        sum+=a[i];    }    return sum;}bool cmp1(const Tnode &x,const Tnode &y){    return x.val<y.val;}bool cmp2(const Tnode &x,const Tnode &y){    return x.num<y.num;}int main(){    int i,preval,temp;    while(scanf("%d",&n),n)    {        for(i=1; i<=n; i++)        {            scanf("%d",&node[i].val);            node[i].num=i;        }        sort(node+1,node+n+1,cmp1);        preval=-1;        for(i=1; i<=n; i++)         // 此循环用来离散化 a[i] 将其范围缩小到maxn内        {                           // 注意这里不能简单地 node[i].val=i;            temp=node[i].val;            if(node[i].val==preval)            {                node[i].val=node[i-1].val;            }            else node[i].val=i;                 preval=temp;        }        sort(node+1,node+n+1,cmp2);        memset(a,0,sizeof(a));        ans=0;        for(i=1; i<=n; i++)        {            ans+=getsum(maxn-5)-getsum(node[i].val);            update(node[i].val);        }        printf("%I64d\n",ans);   //  坑爹 原来逆序数可以很大很大 要用__int64保存  不然会WA    }    return 0;}