Poj 2299 Ultra-QuickSort

1.Link：

http://poj.org/problem?id=2299

2.Content：

Ultra-QuickSort

Time Limit: 7000MS Memory Limit: 65536KTotal Submissions: 41876 Accepted: 15208

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

59105431230

Sample Output

60

Source

Waterloo local 2005.02.05

3.Method:

 

4.Code:

  1 #include<iostream>  2 #include<stdio.h>  3 using namespace std;  4 int a[500002];  5   6 //递归2路归并排序   7 /*void Merge(long long a[],long long b[],int s,int m,int t)  8 {  9      int i=s,j=m+1,k=s; 10      while((i<=m)&&(j<=t)) 11      { 12          if(a[i]<=a[j]) b[k++]=a[i++];   13          else b[k++]=a[j++]; 14      } 15      while(i<=m) b[k++]=a[i++]; 16      while(j<=t) b[k++]=a[j++]; 17 } 18 void MSort(long long a[],long long b[],int s,int t ,int size) 19 { 20      int m; 21      long long c[size]; 22      if(s==t) b[s]=a[t]; 23      else 24      { 25          m=(s+t)/2; 26          MSort(a,c,s,m,size); 27          MSort(a,c,m+1,t,size); 28          Merge(c,b,s,m,t); 29      } 30 } 31 void MergeSort(long long a[],int size) 32 { 33      MSort(a,a,0,size-1,size); 34 }*/ 35  36  37 // 非递归合并排序 38 /*template<class T> 39 void Merge(T a[],T b[],int s,int m,int t) 40 { 41       int i=s,j=m+1,k=s; 42       while(i<=m&&j<=t) 43       { 44          if(a[i]<a[j]) b[k++]=a[i++]; 45          else b[k++]=a[j++]; 46       } 47       while(i<=m) b[k++]=a[i++]; 48       while(j<=t) b[k++]=a[j++];  49 } 50 template<class T> 51 void MergePass(T a[],T b[],int s,int t) 52 { 53       int i; 54       for(i=0;i+2*s<=t;i=i+2*s) 55       { 56         Merge(a,b,i,i+s-1,i+2*s-1); 57      } 58      //剩下的元素个数少于2s 59      if(i+s<t) Merge(a,b,i,i+s-1,t); 60      else for(int j=i;j<=t;j++) b[j]=a[j];  61 } 62 template<class T> 63 void MergeSort(T a[],int n) 64 { 65      T *b=new T[n]; 66      //T b[n]; 67      int s=1; 68      while(s<n) 69      { 70          MergePass(a,b,s,n); 71          s+=s; 72          MergePass(b,a,s,n); 73          s+=s; 74      } 75 }*/ 76  77 long long count=0; 78 //求逆序对 79 void Merge(int a[],int b[],int s,int m,int t) 80 { 81       int i=s,j=m+1,k=s; 82      //int count=0; 83       while(i<=m&&j<=t) 84       { 85          if(a[i]<=a[j]) b[k++]=a[i++]; 86          else 87          { 88              b[k++]=a[j++]; 89              count+=m-i+1; 90          } 91               92       } 93       while(i<=m) b[k++]=a[i++]; 94       while(j<=t) b[k++]=a[j++]; 95       //return count; 96 } 97 void MergePass(int a[],int b[],int s,int t) 98 { 99       int i;100      //int count=0;101       for(i=0;i+2*s<=t;i=i+2*s)102       {103         Merge(a,b,i,i+s-1,i+2*s-1);104      }105      //剩下的元素个数少于2s106      if(i+s<t) Merge(a,b,i,i+s-1,t-1);107      else for(int j=i;j<=t;j++) b[j]=a[j]; 108      //return count;109 }110 void MergeSort(int a[],int n)111 {112      int *b=new int[n];113      //int b[n];114      //int count=0;115      int s=1;116      while(s<n)117      {118          MergePass(a,b,s,n);119          s+=s;120          MergePass(b,a,s,n);121          s+=s;122      }123 }124 125 126 127 int main()128 {129     130     //测试排序正确性 131     /*int a[10];132     for(i=0;i<10;i++) a[i]=10-i;133     for(i=0;i<10;i++) cout<<a[i]<<" ";134     cout<<endl;135     cout<<MergeSort(a,10)<<endl;136     for(i=0;i<10;i++) cout<<a[i]<<" ";*/137 138     int size;139     int i;140     while((cin>>size)&&size!=0)141     {142         count=0;143         for(i=0;i<size;i++)144         {145            scanf("%lld",&a[i]);146         }147         MergeSort(a,size);148         printf("%lld\n",count);149         //测试排序正确性150         //for(i=0;i<size;i++) printf("%lld ",a[i]);151     }152     //system("pause");153     return 1;154 }

 

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