POJ-2299 Ultra-QuickSort（树状数组）（离散化）-白红宇

POJ-2299 Ultra-QuickSort（树状数组）（离散化）

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发布时间：2019-04-30

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Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 56014		Accepted: 20694

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

Source

树状数组求逆序数。由于数组元素最大达到999,999,999，所以不能直接开数组，要先用“离散化”的方法将数组元素映射成1~n，再进行一般的树状数组操作。

#include
   
    #include
    
     #include
     
      #include
      
       using namespace std;typedef long long ll;const int N = 500005;struct node{    int val,pos;}nod[N];int C[N],r[N];int n;bool cmp(node a,node b){    return a.val < b.val;}int lowbit(int t){    return t&(-t);}void add(int t){    while(t <= n)    {        C[t] += 1;        t += lowbit(t);    }}//求得前面小于当前的数的个数int getnum(int t){    int ans = 0;    while(t > 0)    {        ans += C[t];        t -= lowbit(t);    }    return ans;}int main(){    while(~scanf("%d",&n) && n)    {        memset(C,0,sizeof(C));        for(int i = 1;i <= n;i++)        {            scanf("%d",&nod[i].val);            nod[i].pos = i;        }        sort(nod+1, nod+1+n,cmp);//先sort        for(int i = 1;i <= n;i++)//离散化处理            r[nod[i].pos] = i;        ll ans = 0,k;        for(int i = 1;i <= n;i++)        {            add(r[i]);            k = i - getnum(r[i]);            ans += k;        }        printf("%lld\n",ans);    }    return 0;}

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