Ultra-QuickSort(树状数组+离散化)

Ultra-QuickSort  POJ 2299

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 50495		Accepted: 18525

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

5
9
1
0
5
4
3
1
2
3
0

Sample Output

6
0
树状数组维护：c[i]存储比i小的数目.
离散化： 比如输入9 1 0 5 4，将其转化为 5 2 1 4 3，方便存储。
参考别人的代码，看了好半天！

 #include <cstdio>
 #include <iostream>
 #include <algorithm>
 #include <cstring>
 #define LL long long
 #define Max 500000+10
 using namespace std;
 struct node
 {
     int num,order;
 };
 int n;
 int c[Max];
 node in[Max];
 int t[Max];
 int cmp(node a,node b)
 {
     return a.num<b.num;
 }
 void add(int i,int a)
 {
     while(i<=n)
     {
         t[i]+=a;
         i+=i&-i;
     }
 }
 int sum(int i)
 {
     int s=;
     while(i>=)
     {
         s+=t[i];
         i-=i&-i;
     }
     return s;
 }
 int main()
 {
     int i,j;
     freopen("in.txt","r",stdin);
     while(~scanf("%d",&n)&&n)
     {
         memset(t,,sizeof(t));
         for(i=;i<=n;i++)
         {
             scanf("%d",&in[i].num);
             in[i].order=i;
         }
         sort(in+,in++n,cmp);
         for(i=;i<=n;i++)             /*离散化*/
             c[in[i].order]=i;
         LL s=;
         for(i=;i<=n;i++)
         {
             add(c[i],);
             s+=i-sum(c[i]);
     //        cout<<s<<endl;

         }
         printf("%lld\n",s);
     }
 }