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 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这题很简单的样子，就是求冒泡排序的交换次数，but 超时
归并排序，求逆序数，别问我是什么？看着模板写就好
1 #include<iostream> 2 #include<cstring> 3 #include<cstdio> 4 using namespace std; 5 int s[500005],temp[500005]; 6 long long cut;//这里害我WA了一次 7 void merge_sort(int* A,int x,int y,int* T ) 8 { 9 if(y-x>1) 10 { 11 int m=x+(y-x)/2;//划分 12 int p=x,q=m,i=x; 13 merge_sort(A,x,m,T);//递归求解 14 merge_sort(A,m,y,T); 15 while(p<m||q<y) 16 { 17 if(q>=y||(p<m&&A[p]<=A[q])) 18 T[i++]=A[p++];//从左半数组复制到临时空间 19 else 20 { 21 T[i++]=A[q++];//从右半数组复制到临时空间 22 cut+= (m-p);//统计逆序数 23 } 24 } 25 for(i=x; i<y; i++) 26 A[i]=T[i];//从辅助数组复制回原数组 27 } 28 } 29 int main() 30 { 31 int n,i; 32 while(scanf("%d",&n)&&n) 33 { 34 cut=0; 35 for(i=0; i<n; i++) 36 scanf("%d",&s[i]); 37 merge_sort(s,0,n,temp); 38 printf("%lld ",cut); 39 } 40 return 0; 41 }
1 #include<iostream> 2 using namespace std; 3 long long cnt; 4 void merge(int array[],int left,int mid,int right) 5 { 6 int* temp=new int[right-left+1]; 7 int i,j,p; 8 for(i=left,j=mid+1,p=0; i<=mid&&j<=right; p++) 9 { 10 if(array[i]<=array[j])temp[p]=array[i++]; 11 else temp[p]=array[j++],cnt+=(mid-i+1); 12 } 13 while(i<=mid)temp[p++]=array[i++]; 14 while(j<=right)temp[p++]=array[j++]; 15 for(i=left,p=0; i<=right; i++)array[i]=temp[p++]; 16 delete temp; 17 } 18 void mergesort(int array[],int left,int right) 19 { 20 if(left==right)array[left]=array[right]; 21 else 22 { 23 int mid=(left+right)/2; 24 mergesort(array,left,mid); 25 mergesort(array,mid+1,right); 26 merge(array,left,mid,right); 27 } 28 } 29 int main() 30 { 31 int n,array[500005]; 32 while(cin>>n&&n) 33 { 34 cnt=0; 35 for(int i=0; i<n; i++) 36 cin>>array[i]; 37 mergesort(array,0,n-1); 38 cout<<cnt<<endl; 39 } 40 return 0; 41 }
下面这个是网上找的还算好懂得，耗时是我敲得那个的10倍左右