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！
 1 #include <cstdio>
 2 #include <iostream>
 3 #include <algorithm>
 4 #include <cstring>
 5 #define LL long long
 6 #define Max 500000+10
 7 using namespace std; 
 8 struct node
 9 {
10     int num,order;
11 };
12 int n;
13 int c[Max];        
14 node in[Max];
15 int t[Max];
16 int cmp(node a,node b)
17 {
18     return a.num<b.num;
19 }
20 void add(int i,int a)
21 {
22     while(i<=n)
23     {
24         t[i]+=a;
25         i+=i&-i;
26     }
27 }
28 int sum(int i)
29 {
30     int s=0;
31     while(i>=1)
32     {
33         s+=t[i];
34         i-=i&-i;
35     }
36     return s;
37 }
38 int main()
39 {
40     int i,j;
41     freopen("in.txt","r",stdin);
42     while(~scanf("%d",&n)&&n)
43     {
44         memset(t,0,sizeof(t));
45         for(i=1;i<=n;i++)
46         {
47             scanf("%d",&in[i].num);
48             in[i].order=i;
49         }
50         sort(in+1,in+1+n,cmp);
51         for(i=1;i<=n;i++)             /*离散化*/
52             c[in[i].order]=i;
53         LL s=0;
54         for(i=1;i<=n;i++)
55         {
56             add(c[i],1);
57             s+=i-sum(c[i]);
58     //        cout<<s<<endl;
59             
60         }
61         printf("%lld\n",s);
62     }
63 }

 

posted @ 2015-11-26 12:56  御心飞行  阅读(261)  评论(0)  编辑  收藏  举报