PAT (Advanced Level) Practise - 1098. Insertion or Heap Sort (25)

 http://www.patest.cn/contests/pat-a-practise/1098

According to Wikipedia:

Insertion sort iterates, consuming one input element each repetition, and growing a sorted output list. Each iteration, insertion sort removes one element from the input data, finds the location it belongs within the sorted list, and inserts it there. It repeats until no input elements remain.

Heap sort divides its input into a sorted and an unsorted region, and it iteratively shrinks the unsorted region by extracting the largest element and moving that to the sorted region. it involves the use of a heap data structure rather than a linear-time search to find the maximum.

Now given the initial sequence of integers, together with a sequence which is a result of several iterations of some sorting method, can you tell which sorting method we are using?

Input Specification:

Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then in the next line, N integers are given as the initial sequence. The last line contains the partially sorted sequence of the N numbers. It is assumed that the target sequence is always ascending. All the numbers in a line are separated by a space.

Output Specification:

For each test case, print in the first line either "Insertion Sort" or "Heap Sort" to indicate the method used to obtain the partial result. Then run this method for one more iteration and output in the second line the resuling sequence. It is guaranteed that the answer is unique for each test case. All the numbers in a line must be separated by a space, and there must be no extra space at the end of the line.

Sample Input 1:

10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0

Sample Output 1:

Insertion Sort
1 2 3 5 7 8 9 4 6 0

Sample Input 2:

10
3 1 2 8 7 5 9 4 6 0
6 4 5 1 0 3 2 7 8 9

Sample Output 2:

Heap Sort
5 4 3 1 0 2 6 7 8 9



此题是2015年春季的研究生入学考试复试时的机试题，链接 http://www.patest.cn/contests/graduate-entrance-exam-2015-03-20

#include<cstdio>
#include<cstring> 

void headshift(int *a,int i,int len)//数组  要处理的节点下标  当前处理的堆的长度 
{
     if(i>=len/2 || i<0) return;//叶节点可看作一个满足性质的堆，所以只对非叶子结点进行处理即可 
     int max=i; //调整一个节点为（节点值、左孩子节点值、右孩子节点值）三者中的最大值 
     if(2*i+1<=len-1 && a[max]<a[2*i+1]) max=2*i+1;
     if(2*i+2<=len-1 && a[max]<a[2*i+2]) max=2*i+2;
     if(max!=i) 
     {
         int temp=a[max];
         a[max]=a[i],a[i]=temp;
         headshift(a,max,len);//某节点与其某个孩子节点的数值交换调整后，可能会影响该孩子节点的堆的性质，所以要向下调整 
     }
     return;
}

int match(int *temp,int *a,int len)
{
    for(int i=0;i<len;i++)
        if(temp[i]!=a[i]) return 0;   
    return 1;
}

void print(int *a,int len)
{     
     for(int i=0;i<len;i++)
     {
             if(i) printf(" %d",a[i]);
             else printf("%d",a[i]);
     }
     return;
}

void Insertion(int *a,int i)
{
     int loc=i-1,temp=a[i];
     for(;loc>=0;loc--) if(a[loc]<=temp) break;//寻找应该插入的位置 
     loc++;
     if(loc!=i) for(int j=i;j>loc;j--) a[j]=a[j-1];//比当前处理值大的元素相应后移 
     a[loc]=temp;
}
int main()
{
    
   int num=0,data[100]={0},sorttemp[100]={0},a[100]={0};
   scanf("%d",&num);
   for(int i=0;i<num;i++) scanf("%d",data+i);
   for(int i=0;i<num;i++) scanf("%d",sorttemp+i);   
   
   for(int i=0;i<num;i++) a[i]=data[i];
   for(int i=1;i<num;i++) //i=0时，不构成插入排序的定义 因为0元素之前没有有序子序列 
   {
           Insertion(a,i); 
           if(match(sorttemp,a,num))//if（match） 再处理一轮+输出+return 
           {
                   if(i<num-1) i++;  //i++易漏掉  one more iteration
                   Insertion(a,i);
                   printf("Insertion Sort\n");
                   print(a,num);
                   return 0; 
           }
   }
    
   
   for(int i=0;i<num;i++) a[i]=data[i];
   int len=num,temp=0;   
  
   for(int i=num/2-1;i>=0;i--) //构建大根堆
           headshift(a,i,len);           
  
   for(int i=num-1;i>=0;i--) //排序             
   {
           temp=a[0],a[0]=a[i],a[i]=temp;//顶尾交换  表示当前最大值已处理    
           len--;//堆长度-1
           headshift(a,0,len);//headshift新顶
           if(match(sorttemp,a,num))//if（match） 再处理一轮+输出+return 
           {
                   i--; //i-- 易漏掉  one more iteration
                   temp=a[0],a[0]=a[i],a[i]=temp;
                   len--;
                   headshift(a,0,len);
                   printf("Heap Sort\n");
                   print(a,num);
                   return 0;
           }
   }
    return 0;
}