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和前一题差不多，把归并排序换成了堆排序。要点还是每一次排序进行判断
开始犯了个错误 堆排序该用origin2 结果一直在排序origin ，误导了半天以为是逻辑错误。。。一直在检查逻辑
建立最大堆
排序并调整下滤
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?
Each input file contains one test case. For each case, the first line gives a positive integer N (≤100). Then in the next line, NN 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.
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.
10
3 1 2 8 7 5 9 4 6 0
1 2 3 7 8 5 9 4 6 0
Insertion Sort
1 2 3 5 7 8 9 4 6 0
10
3 1 2 8 7 5 9 4 6 0
6 4 5 1 0 3 2 7 8 9
Heap Sort
5 4 3 1 0 2 6 7 8 9
1 #include <iostream> 2 #include <cstdio> 3 #include <cstdlib> 4 using namespace std; 5 6 typedef int ElementType; 7 bool Judge(int origin[], int changed[], int len) 8 { 9 for(int i = 0; i < len; i++) { 10 if(origin[i] != changed[i]) 11 return false; 12 } 13 return true; 14 } 15 16 //插入排序 (需排序数组，数组长度，排序次数) 17 void isInsertionSort(ElementType origin[], int N, int times) 18 { 19 int i; 20 ElementType temp = origin[times]; //取出未排序序列中的第一个元素 21 for (i = times; i > 0 && origin[i-1] > temp; i-- ) 22 origin[i] = origin[i-1]; //依次与已排序序列中元素比较并右移 23 origin[i] = temp; 24 } 25 26 void Swap( ElementType *a, ElementType *b ) 27 { 28 ElementType t = *a; 29 *a = *b; 30 *b = t; 31 } 32 /* 改编PercDown( MaxHeap H, int p )*/ 33 void PercDown( ElementType A[], int p, int N ) 34 { 35 /* 将N个元素的数组中以A[p]为根的子堆调整为最大堆 */ 36 int Parent, Child; 37 38 ElementType X = A[p]; /* 取出根结点存放的值 */ 39 for( Parent=p; (Parent*2+1) < N; Parent=Child ) { 40 Child = Parent * 2 + 1; 41 if( (Child != N-1) && (A[Child] < A[Child+1]) ) 42 Child++; /* Child指向左右子结点的较大者 */ 43 if( X >= A[Child] ) break; /* 找到了合适位置 */ 44 else /* 下滤X */ 45 A[Parent] = A[Child]; 46 } 47 A[Parent] = X; 48 } 49 50 void HeapSort(ElementType A[], int N, int changed[]) 51 { 52 for(int i = N/2-1; i >= 0; i--) /* 建立最大堆 */ 53 PercDown( A, i, N ); 54 55 for(int i = N-1; i > 0; i--) { 56 /* 删除最大堆顶 */ 57 Swap(&A[0], &A[i] ); 58 PercDown(A, 0, i); 59 60 if( Judge(A, changed, N) ) { //是堆排序 61 printf("Heap Sort\n"); 62 63 Swap(&A[0], &A[i-1] ); //在执行一次堆排序 64 PercDown(A, 0, i-1); 65 for(int j = 0; j < N-1; j++) 66 printf("%d ",A[j]); 67 printf("%d\n",A[N-1]); 68 return; 69 } 70 } 71 } 72 73 int main() 74 { 75 int N; 76 int origin[105],origin2[105],changed[105]; 77 scanf("%d", &N); 78 for(int i = 0; i < N; i++) { //origin origin1 初始序列 79 scanf("%d",&origin[i]); 80 origin2[i] = origin[i]; 81 } 82 for(int i = 0; i < N; i++) //changed 排序后序列 83 scanf("%d",&changed[i]); 84 85 for(int i = 1; i < N; i++) { 86 isInsertionSort(origin, N, i); 87 if( Judge(origin, changed,N) ) { //是插入排序 88 printf("Insertion Sort\n"); 89 isInsertionSort(origin, N, i+1); 90 for(int j = 0; j < N-1; j++) 91 printf("%d ",origin[j]); 92 printf("%d\n",origin[N-1]); 93 return 0; 94 } 95 } 96 97 HeapSort(origin2,N,changed); 98 return 0; 99 }
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原文地址：http://www.cnblogs.com/kuotian/p/5461209.html
