poj 2299

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 70489		Accepted: 26437

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

Source

 

 

 

 //求逆序对

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <cstring>
 4 #include <string>
 5 #include <algorithm>
 6 using namespace std;
 7 #define ll long long 
 8 #define  N 500009
 9 #define  gep(i,a,b)  for(ll  i=a;i<=b;i++) 
10 #define  mem(a,b)  memset(a,b,sizeof(a))
11 #define  lowbit(x)  x&(-x)
12 ll  c[N],a[N],n;//一定要用ll
13 struct Node{
14     ll  id,val;
15 }nod[N];
16 void update(ll i,ll num)
17 {
18     while(i<=n){
19         c[i]+=num;
20         i+=lowbit(i);
21     }
22 }
23 ll getsum(ll  n)
24 {
25     ll sum=0;
26     while(n>0){
27         sum+=c[n];
28         n-=lowbit(n);
29     }
30     return sum;
31 }
32 bool cmp(Node a,Node b)
33 {
34     return a.val<b.val;
35 }
36 int main()
37 {
38     while(~scanf("%lld",&n)&&n){
39         mem(a,0);
40         mem(c,0);
41         gep(i,1,n){
42             scanf("%lld",&nod[i].val);
43             nod[i].id=i;
44         }
45         sort(nod+1,nod+1+n,cmp);//要先排序           
46         int x=1;
            a[nod[1].id]=x;
47         gep(i,2,n)
48         {
49             
50             if(nod[i].val!=nod[i-1].val){//离散化
51                x++;
52             }            
                 a[nod[i].id]=x;
53         }    
54         ll ans=0;
55         gep(i,1,n){
56             update(a[i],1);
57             ans+=getsum(n)-getsum(a[i]);//左边比我大的数的数目
58         }
59         printf("%lld\n",ans);
60     }
61     return 0;
62 }