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poj 2299 Ultra-QuickSort(树状数组)

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
题意 :多组输入,输入n,n==0结束,接着输入n个数。把输入的数排好序需要几步。
思路:由于输入的数不是从1开始公差为1的递增序列,先用两个快排转换一下,代入树状数组就ok了!
ac代码:
#include<stdio.h>#include<iostream>#include<string.h>#include<algorithm>using namespace std;long long d1[5000000];int n;struct node                                      //g存数,h存下标{    int g,h;} s[5000000];int cmp(struct node a,struct node b)             //控制排序的方向从大到小{    return a.g>b.g;}int cmp1(struct node a,struct node b)            //控制排序方向从小到大{    return a.h<b.h;}int lowbit(int x)                               {    return x&(-x);}long long sum(int x)                             //树状数组求和,这题是求x前面有多少个数,因为把所有数转换成公差为1的递增序列,求出的和就是答案{    long long res=0;    while(x>0)    {        res+=d1[x];        x-=lowbit(x);    }    return res;}void add(int x)                                  //更新树状数组{    while(x<=n)    {        d1[x]++;        x+=lowbit(x);    }}int main(){    int a,b,i,j;    long long num;    while(scanf("%d",&n)!=EOF)    {        if(n==0)        break;        num=0;        memset(d1,0,sizeof(d1));        for(i=1; i<=n; i++)        {            scanf("%d",&s[i].g);            s[i].h=i;        }        sort(s+1,s+n+1,cmp);                          //按g值的大小排序        for(i=1; i<=n; i++)            s[i].g=i;        sort(s+1,s+n+1,cmp1);                         //按h排序 ,就是原来的下标        for(i=1;i<=n;i++)        {            num+=sum(s[i].g);                         //把这个数前面有的数的个数相加,就是答案            add(s[i].g);        }        printf("%lld\n",num);    }}
 

poj 2299 Ultra-QuickSort(树状数组)