Poj 2299 Ultra-QuickSort

1.Link：

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

2.Content：

Ultra-QuickSort

Time Limit: 7000MS		Memory Limit: 65536K
Total 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

5

9

1

0

5

4

3

1

2

3

0

Sample Output

6

0

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2s

106      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 }

 

5.Reference: