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NUC_TeamTEST_C && POJ2299(只有归并)

时间:2014-10-27 00:17:11      阅读:243      评论:0      收藏:0      [点我收藏+]

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Ultra-QuickSort
Time Limit: 7000MS   Memory Limit: 65536K
Total Submissions: 42627   Accepted: 15507

Description

bubuko.com,布布扣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

Source

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 }

 

NUC_TeamTEST_C && POJ2299(只有归并)

标签:des   style   blog   http   color   io   os   ar   for   

原文地址:http://www.cnblogs.com/ya-cpp/p/4052965.html

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