PAT Advanced 1155 Heap Paths (30分)
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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))
One thing for sure is that all the keys along any path from the root to a leaf in a max/min heap must be in non-increasing/non-decreasing order.
Your job is to check every path in a given complete binary tree, in order to tell if it is a heap or not.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (1), the number of keys in the tree. Then the next line 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, first print all the paths from the root to the leaves. Each path occupies a line, with all the numbers separated by a space, and no extra space at the beginning or the end of the line. The paths must be printed in the following order: for each node in the tree, all the paths in its right subtree must be printed before those in its left subtree.
Finally 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.
Sample Input 1:
8
98 72 86 60 65 12 23 50
Sample Output 1:
98 86 23
98 86 12
98 72 65
98 72 60 50
Max Heap
Sample Input 2:
8
8 38 25 58 52 82 70 60
Sample Output 2:
8 25 70
8 25 82
8 38 52
8 38 58 60
Min Heap
Sample Input 3:
8
10 28 15 12 34 9 8 56
Sample Output 3:
10 15 8
10 15 9
10 28 34
10 28 12 56
Not Heap
题目大意:给定一个堆数组,判定是大顶堆,小顶堆,还是不是堆。并且还需要打印自上而下,自左向右的顺序
解决方案:使用一个vector进行存储,进行DFS进行遍历,当为叶子节点的时候,记录路径(不能立即打印,否则是反的),用stack进行记录路径顺序,最后打印。
解决判定顶堆,采用一个int值进行记录,如果是1则是大顶堆,0不是顶堆,-1小顶堆。紧接着,我们每次DFS进行判定最后两个数据,如果是符合顶堆的路径,则不需要管。否则需要进行赋值heapJudge为0
#include <iostream> #include <vector> #include <stack> using namespace std; vector<int> v; vector<int> path; stack<string> res; /** 判定是否为堆 */ int heapJudge;// -1 minHeap 0 notHeap 1 maxHeap void DFS(int index){ /** 每次DFS的时候,最后两个path数据进行比较,如果不符合顶堆条件,则进行置0 */ if(path.size()>1){ if(heapJudge==1&&path[path.size()-1]>path[path.size()-2]) heapJudge=0; if(heapJudge==-1&&path[path.size()-1]<path[path.size()-2]) heapJudge=0; } /** 如果大于数据的长度,则进行push到栈里面,稍后打印 */ if(index>=v.size()){ string str=""; for(int i=0;i<path.size();i++){ if(i!=0) str+=" "; str+=to_string(path[i]); } str+=" "; res.push(str); return ; } /** push进一个数据 */ path.push_back(v[index]); /** 进行DFS */ DFS(index*2); /** 如果有右子树,还要进行打印遍历,否则不需要,因为如果是叶子,会打印2次*/ if(index*2+1<v.size())DFS(index*2+1); path.pop_back(); } int main(){ // 进行先序打印 int N;cin>>N; v.resize(N+1); for(int i=1;i<=N;i++) cin>>v[i]; /** 判断第一个数据和第二个数据,如果大则是大顶堆,否则是小顶堆 */ v[1]>v[2] ? heapJudge=1:heapJudge=-1; DFS(1); while(!res.empty()){ cout<<res.top(); res.pop(); } if(heapJudge==1) printf("Max Heap"); else if(heapJudge==0) printf("Not Heap"); else printf("Min Heap"); system("pause"); return 0; }
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