A.归并排序

 

归并排序 （求逆序数）

归并排序：递归+合并+排序

时间复杂度：O（n logn）    空间复杂度：O（n）

用途：1.排序  2.求逆序对数

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.       

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

题意：
给出长度为n的无序序列，每次只能交换相邻的两个数，求至少要交换多少次才使得该序列为递增序列。

分析：
1.根据时间范围估计时间复杂度，归并排序的时间复杂度一般为O（n logn）
2.算法步骤： 1.划分问题：把序列分为元素个数尽量相等的两半
           2.递归求解：把两半元素分别排序
           3.合并问题：把两个有序表合成一个
3.保存时，用long long 类型，否则会wa的

代码：
 1 #include<cstdio>
 2 #include<iostream>
 3 using namespace std;
 4 const int maxn=500001;
 5 
 6 int a[maxn],b[maxn];
 7 long long ans;
 8 
 9 void merge(int l,int m,int r)       //归并排序
10 {
11     int i=l,j=m+1,t=0;
12     while(i<=m&&j<=r)   //从小到大排序
13     {
14         if(a[i]>a[j])
15         {
16             b[t++]=a[j++];
17             ans+=m-i+1;
18         }
19         else b[t++]=a[i++];
20     }
21     while(i<=m)
22         b[t++]=a[i++];
23     while(j<=r)
24         b[t++]=a[j++];
25     for(int i=0;i<t;i++)   //合并
26         a[l+i]=b[i];
27 }
28 
29 void merge_sort(int l,int r)  //划分问题
30 {
31     if(l<r)
32     {
33         int m=(l+r)/2;
34         merge_sort(l,m);
35         merge_sort(m+1,r);
36         merge(l,m,r);
37     }
38 }
39 
40 int main()
41 {
42     int n;
43     while(scanf("%d",&n)!=EOF)
44     {
45         if(n==0)
46             break;
47         ans=0;
48         for(int i=0;i<n;i++)
49             scanf("%d",&a[i]);
50         merge_sort(0,n-1);
51         printf("%lld\n",ans);
52     }
53     return 0;
54 }
View Code

本来想着完全不看别人的代码都由自己做，可是看了一天了只是有了思路，还是写不出来。。。又看来了别人的代码。。。

posted @ 2015-08-04 20:39  tmj  阅读(236)  评论(0)  编辑  收藏  举报