Poj 2299 Ultra-QuickSort

1.Link：

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:

posted on 2014-08-24 21:01 mobileliker 阅读(157) 评论(0) 编辑收藏举报

Time Limit: 7000MS		Memory Limit: 65536K
Total Submissions: 41876		Accepted: 15208