题目网址:http://acm.hust.edu.cn/vjudge/contest/view.action?cid=109331#problem/A
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.
Ultra-QuickSort produces the output
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:输入n个数,相邻两个数可以交换位置进行从小到大排序,求最小交换次数。
解题思路:用线段树的思想将数组一半一半的划分成小的区间,最后划分为只有两个数的区间,这两个数进行比较交换位置,记录交换次数就。先进行前两个数的比较,然后扩大区间为(left,right),
可知(left,right)区间中,(left,center)和(center+1,right)区间里的数已经排好序,将(center+1,right)中的数一个一个与(left,center)中的数比较大小,用另一个数组记录
新的排序后的位置(相当于向前插入数),并记录交换次数。
#include <iostream> #include <algorithm> #include <cstring> #include <cstdio> #include <cstdlib> using namespace std; int a[500010]; long long num; int temp[500010]; void Merge(int arr[], int left, int center, int right) { int i=left; int j=center+1; int k=0; while (i<=center&&j<=right) { ///可知(left,right)区间中,(left,center)和(center+1,right)区间里的数已经排好序 if (arr[i] > arr[j]) { temp[k++] = arr[j++];/// num+= center+1-i; } else temp[k++] = arr[i++];///记录小的数,以便得到排序后新的序列; } while (i <= center) /// temp[k++] = arr[i++]; while (j <= right) ///这个和上个while语句两个while语句记录了原数组交换次序后的新的数组; temp[k++] = arr[j++]; for (i = left, k = 0; i <= right; i++, k++)///将array[]数组变为新的数组; arr[i] = temp[k]; } void mergeSort(int arr[], int left, int right) { if (left<right) { int center = (left + right) / 2; mergeSort(arr, left, center); /// mergeSort(arr, center + 1, right);///将数组划分为两个区间; Merge(arr, left, center, right); ///将这个区间里的数进行排序; } } int main() { int n; while(scanf("%d",&n)&&n) { num=0; for(int i=0;i<n;i++) scanf("%d",&a[i]); mergeSort(a,0,n-1); printf("%lld ",num); } }