pat 1147
1147 Heaps (30分)
In computer science, a heap is a specialized tree-based data structure that satisfies the heap property: if P is a parent node of C, then the key (the value) of P is either greater than or equal to (in a max heap) or less than or equal to (in a min heap) the key of C. A common implementation of a heap is the binary heap, in which the tree is a complete binary tree. (Quoted from Wikipedia at https://en.wikipedia.org/wiki/Heap_(data_structure))
Your job is to tell if a given complete binary tree is a heap.
Input Specification:
Each input file contains one test case. For each case, the first line gives two positive integers: M (≤ 100), the number of trees to be tested; and N (1 < N ≤ 1,000), the number of keys in each tree, respectively. Then M lines follow, each contains N distinct integer keys (all in the range of int), which gives the level order traversal sequence of a complete binary tree.
Output Specification:
For each given tree, print in a line Max Heap
if it is a max heap, or Min Heap
for a min heap, or Not Heap
if it is not a heap at all. Then in the next line print the tree's postorder traversal sequence. All the numbers are separated by a space, and there must no extra space at the beginning or the end of the line.
Sample Input:
3 8 98 72 86 60 65 12 23 50 8 38 25 58 52 82 70 60 10 28 15 12 34 9 8 56
Sample Output:
Max Heap 50 60 65 72 12 23 86 98 Min Heap 60 58 52 38 82 70 25 8 Not Heap 56 12 34 28 9 8 15 10
题意:给定一个完全二叉树,要求判断是否是堆,并输出这个完全二叉树的后序遍历序列
思路:数组存放完全二叉树,判断堆只需一次检查每个元素是否满足堆的条件。后序遍历序列参考树的四种遍历方法(递归实现)
代码如下:
#include<cstdio> #include<vector> using namespace std; vector<int> v; void check(){ int max=1; int min=1; for(int i=1;i<=v.size();i++){ if(max==0&&min==0){ printf("Not Heap\n"); return; } if(2*i<v.size()){ if(v[i]>=v[i*2]) min=0; if(v[i]<=v[i*2]) max=0; } if(2*i+1<v.size()){ if(v[i]>=v[i*2+1]) min=0; if(v[i]<=v[i*2+1]) max=0; } } if(max==1) printf("Max Heap\n"); if(min==1) printf("Min Heap\n"); } void postOrder(int i){ if(i<v.size()){ if(2*i<v.size()) postOrder(2*i); if(2*i+1<v.size()) postOrder(2*i+1); if(i==1) printf("%d",v[i]); else printf("%d ",v[i]); } } int main(){ int n,m; scanf("%d%d",&n,&m); v.resize(m+1); for(int i=0;i<n;i++){ v.clear(); v.resize(m+1); for(int j=1;j<=m;j++){ scanf("%d",&v[j]); } check(); postOrder(1); printf("\n"); } return 0; }