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Ultra-QuickSort
Description
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 |
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 这道题就是通过求逆序数的和，来求出序列所有的交换次数。用冒泡排序直接会TLE（相当于暴力了），
这道题有3种解法，树状数组、线段树、归并排序（O（N*lgN））
其中归并排序的写法应该是最简单的，树状数组、线段树要用到离散化，博客后续还会跟上
归并写法，其实归并写法，自己并不是很熟练，推介大牛博客
http://blog.163.com/zhaohai_1988/blog/static/20951008520127321239701/
大牛代码真心漂亮!!中间核心，就是学大牛写的
1 #include <cstdio>
2 #include <iostream>
3 #include <cstring>
4 #define LL long long //LL 代替 long long 的写法 中间数据会超出 int
5 using namespace std;
6
7 const int max_size = 500010;
8
9 int arry[max_size], tmp_arry[max_size];
10
11 LL Merge(int *arr, LL beg, LL mid, LL end, int *tmp_arr)
12 {
13 memcpy(tmp_arr+beg, arr+beg, sizeof(int)*(end-beg+1));
14 LL i = beg;
15 LL j = mid + 1;
16 LL k = beg;
17 LL inversion = 0;
18 while(i <= mid && j <= end)
19 {
20 if(tmp_arr[i] <= tmp_arr[j]) ///如果合并逆序数的时候，前边小于等于后边，就不用记录逆序数的值
21 {
22 arr[k++] = tmp_arr[i++];
23 }else{
24 arr[k++] = tmp_arr[j++]; ///如果不是，则要记录逆序数的值
25 inversion += (mid - i + 1);///简单画下就能看出
26 }
27 }
28
29 while(i <= mid) ///把没有并入arr数组的数并入
30
31 {
32 arr[k++] = tmp_arr[i++];
33 }
34 while(j <= end)
35 {
36 arr[k++] = tmp_arr[j++];
37 }
38 return inversion;
39 }
40
41 LL MergeInversion(int *arr, LL beg, LL end, int *tmp_arr)
42 {
43 LL inversions = 0;
44 if(beg < end)
45 {
46 LL mid = (beg + end) >> 1;
47 inversions += MergeInversion(arr, beg, mid, tmp_arr); ///分成两段分别进行记录，递归的进行下去，找逆序数和
48 inversions += MergeInversion(arr, mid+1, end, tmp_arr);
49 inversions += Merge(arr, beg, mid, end, tmp_arr);
50 }
51 return inversions;
52 }
53
54 int main()
55 {
56 LL n;
57
58 while(cin >> n)
59 {
60 if(n == 0)
61 break;
62 for(int i = 0; i < n; ++i)
63 scanf("%d", &arry[i]);
64 memcpy(tmp_arry, arry, sizeof(int)*n);
65 cout << MergeInversion(arry, 0, n-1, tmp_arry) << endl;
66 }
67 return 0;
68 }