Ultra-QuickSort
Time Limit: 7000MS
Memory Limit: 65536K
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
：：本题其实就是要求出有多少逆序对。本题还要虚拟化，因为0<=a[i]<=999,999,999，开一个数组大小为1,000,000,000*4铁定超内存
1: #include <iostream>
2: #include <cstdio>
3: #include <cstring>
4: #include <algorithm>
5: using namespace std;
6: #define lson l,m,rt<<1
7: #define rson m+1,r,rt<<1|1
8: typedef long long ll;
9: const int maxn=555555;
10: int col[maxn<<2];
11: int a[maxn],b[maxn],n;
12:
13: void build(int l,int r,int rt)
14: {
15: col[rt]=0;
16: if(l==r) return ;
17: int m=(r+l)>>1;
18: build(lson);
19: build(rson);
20: }
21:
22: int find(int x)
23: {
24: int l=1,r=n;
25: while(l<=r)
26: {
27: int m=(l+r)>>1;
28: if(x==a[m]) return m;
29: if(x>a[m]) l=m+1;
30: else r=m-1;
31: }
32: return 0;
33: }
34:
35: void update(int p,int l,int r,int rt)
36: {
37: col[rt]++;
38: if(l==r) return ;
39: int m=(l+r)>>1;
40: if(p<=m) update(p,lson);
41: else update(p,rson);
42: }
43:
44: int query(int p,int l,int r,int rt)
45: {
46: if(p<=l) return col[rt];
47: int m=(l+r)>>1;
48: int ans=0;
49: if(p<=m)
50: ans=col[rt<<1|1]+query(p,lson);
51: else
52: ans=query(p,rson);
53: return ans;
54: }
55:
56: int main()
57: {
58: int i;
59: while(scanf("%d",&n),n)
60: {
61: build(1,n,1);
62: for(i=1; i<=n; i++)
63: {
64: scanf("%d",a+i);
65: b[i]=a[i];
66: }
67: sort(a+1,a+n+1);//让数组a升序排序，那么等下b就可以通过a来求出对应的树是第几大（这算是虚拟化吧）
68: ll sum=0;
69: for(i=1; i<=n; i++)
70: {
71: int t=find(b[i]);
72: sum+=(ll)query(t,1,n,1);
73: update(t,1,n,1);
74: }
75: printf("%lld\n",sum);
76: }
77: return 0;
78: }