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
题目大意:
给出长度为n的序列,每次只能交换相邻的两个元素,问至少要交换几次才使得该序列为递增序列。
刚刚学了时间复杂度, 用归并排序Mergesort了,O(nlogn),省时,不会超时。
这里用归并排序并不是为了求交换次数,而是为了求序列的逆序数,而一个乱序序列的逆序数 = 在只允许相邻两个元素交换的条件下,得到有序序列的交换次数。
案例中的
9 1 0 5 4
要把它排列为升序0,1,4,5,9
而对于序列9 1 0 5 4
9后面却有4个比9小的元素,因此9的逆序数为4
1后面只有1个比1小的元素0,因此1的逆序数为1
0后面不存在比他小的元素,因此0的逆序数为0
5后面存在1个比他小的元素4, 因此5的逆序数为1
4是序列的最后元素,逆序数为0
因此序列9 1 0 5 4的逆序数 t=4+1+0+1+0 = 6 ,就是交换次数
(自己还不是很理解,所以拐到同学的解释来解释了........)
代码如下:
1 #include <stdio.h> 2 int n,A[500005],T[500005],i; 3 long long ans=0; 4 void merge_sort(int* A,int x,int y,int*T){ 5 if(y-x>1){ 6 int m=x+(y-x)/2; 7 int p=x,q=m,i=x; 8 merge_sort(A,x,m,T); 9 merge_sort(A,m,y,T); 10 while(p<m||q<y){ 11 if(q>=y||(p<m&&A[p]<=A[q])){ 12 T[i++]=A[p++]; 13 } 14 else { 15 T[i++]=A[q++]; 16 ans+=m-p; 17 } 18 } 19 for(i=x;i<y;i++) A[i]=T[i]; 20 } 21 } 22 int main() 23 { 24 while(scanf("%d",&n)==1&&n){ 25 ans=0; 26 for(int i=0;i<n;i++) 27 scanf("%d",&A[i]); 28 merge_sort(A,0,n,T); 29 printf("%lld ",ans); 30 } 31 return 0; 32 }
注意:保存逆序数时,必须要用long long型定义,否则会WA的。。。