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 }