POJ 2299 Ultra-QuickSort
离散化+树状数组求逆序数
9 1 0 5 4
3
1 2 3
0
0
* This source code was highlighted by YcdoiT. ( style: Codeblocks )
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 35024 | Accepted: 12608 |
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.
Ultra-QuickSort produces the output
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 ≤ 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
59 1 0 5 4
3
1 2 3
0
Sample Output
60
Source
#include <iostream> #include <cstdio> #include <cstring> #include <algorithm> using namespace std; struct node { int v,ord; }s[550000]; bool cmp(node a,node b) { return a.v<b.v; } int ar[550000],tree[550000],n; int lowbit(int x) { return x&-x; } void update(int x,int v) { while(x<=n) { tree[x]+=v; x+=lowbit(x); } } int getsum(int x) { int sum=0; while(x>0) { sum+=tree[x]; x-=lowbit(x); } return sum; } int main() { while(scanf("%d",&n)&&n) { for(int i=1;i<=n;i++) { scanf("%d",&s s } sort(s+1,s+n+1,cmp); for(int i=1;i<=n;i++) ar[s memset(tree,0,sizeof(tree)); long long int ans=0; for(int i=1;i<=n;i++) { update(ar ans+=i-getsum(ar } printf("%I64d\n",ans); } return 0; } |