poj 2299 Ultra-QuickSort (树状数组)
Ultra-QuickSort
Time Limit: 7000MS | Memory Limit: 65536K | |
Total Submissions: 37971 | Accepted: 13672 |
Description
![](http://poj.org/images/2299_1.jpg)
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
1 //Accepted 9356K 563MS C++ 1145B 2 /* 3 4 题意: 5 给出一串数列,问用冒泡排序发要进行多少次交换。 6 7 树状数组: 8 我的理解是这题和poj 2352类似,不过这题一定要加上离散化。 9 10 这题就是要求每个元素中排在其前面且比它大的元素的数量的和。 11 我转换成和之前那题一样的思想,由最大的数减去该数的排位, 12 最后求出来的a[i]就是所求,表示前面有i个比它大的数。然后每个 13 数都要交换i次,故要乘上i。 14 15 16 */ 17 #include<stdio.h> 18 #include<stdlib.h> 19 #include<iostream> 20 #define N 500005 21 using namespace std; 22 struct node{ 23 int val; 24 int pos; 25 }; 26 node no[N]; 27 int b[N],a[N],c[N]; 28 int cmp(const void*a,const void*b) 29 { 30 return (*(node*)a).val-(*(node*)b).val; 31 } 32 inline int lowbit(int i) 33 { 34 return i&(-i); 35 } 36 void update(int k,int detal) 37 { 38 for(;k<N;k+=lowbit(k)) 39 c[k]+=detal; 40 } 41 inline int getsum(int k) 42 { 43 int t=0; 44 for(;k>0;k-=lowbit(k)) 45 t+=c[k]; 46 return t; 47 } 48 int main(void) 49 { 50 int n; 51 while(scanf("%d",&n)!=EOF && n) 52 { 53 __int64 sum=0; 54 for(int i=0;i<n;i++){ 55 scanf("%d",&no[i].val); 56 no[i].pos=i; 57 } 58 qsort(no,n,sizeof(no[0]),cmp); 59 for(int i=0;i<n;i++) b[no[i].pos]=i; 60 //for(int i=0;i<n;i++) printf("%d\n",b[i]); 61 memset(c,0,sizeof(c)); 62 memset(a,0,sizeof(a)); 63 for(int i=0;i<n;i++){ 64 int t=N-b[i]-5; 65 a[getsum(t)]++; 66 update(t,1); 67 } 68 for(int i=1;i<n;i++) sum+=i*a[i]; 69 printf("%I64d\n",sum); 70 } 71 return 0; 72 }