Ultra-QuickSort
| Time Limit: 7000MS | Memory Limit: 65536K | |
| Total Submissions: 54883 | Accepted: 20184 |
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 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[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
题意：给你长度为n个不同的数，问最少需要交换多少次使得升序 只能交换相邻的两个 其实就是冒泡排序的过程
题解：直接模拟的话会超时 树状数组处理，n只有100000 但是0 ≤ a[i] ≤ 999,999,999
如果开树状数组的话 会炸 所以先离散化一下 只要记录数的相对大小就可以了。求每个数的逆序数对数然后求和输出
逆序对：设A[1..n]是一个包含N个非负整数的数组。如果在i〈 j的情况下，有A〉A[j]，则(i,j)就称为A中的一个逆序对。
还有一种归并排序的写法 再补
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 //#include<bits/stdc++.h> 8 #include<iostream> 9 #include<cstring> 10 #include<cstdio> 11 #include<map> 12 #include<algorithm> 13 #include<queue> 14 #include<cmath> 15 #define ll __int64 16 #define PI acos(-1.0) 17 #define mod 1000000007 18 using namespace std; 19 int n; 20 struct node 21 { 22 int v; 23 int pos; 24 }N[500005]; 25 int a[500005]; 26 int tree[500005]; 27 bool cmp(struct node aa,struct node bb) 28 { 29 return aa.v<bb.v; 30 } 31 int lowbit(int t) {return t&(-t);} 32 void add(int i,int v){ 33 for(;i<=n;i+=lowbit(i)) 34 tree[i]+=v; 35 } 36 int sum(int i){//sum 就是统计之前有多少个小于i的数 37 int ans=0; 38 for(;i>0;i-=lowbit(i)) 39 ans+=tree[i]; 40 return ans; 41 } 42 int main() 43 { 44 while(~scanf("%d",&n)){ 45 if(n==0) 46 break; 47 for(int i=1;i<=n;i++) 48 { 49 scanf("%d",&N[i].v); 50 N[i].pos=i; 51 } 52 sort(N+1,N+1+n,cmp); 53 for(int i=1;i<=n;i++)//离散化 54 a[N[i].pos]=i; 55 for(int i=1;i<=n;i++)//初始化 56 tree[i]=0; 57 ll ans=0; 58 for(int i=1;i<=n;i++) 59 { 60 add(a[i],1);//注意修改值为1 61 ans+=(i-sum(a[i])); 62 } 63 printf("%I64d ",ans); 64 } 65 return 0; 66 }