• 归并排序


    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。。。 

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  • 原文地址:https://www.cnblogs.com/huangguodong/p/4699634.html
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