PAT Advanced 1155 Heap Paths (30分)

2020-01-17 19:01:53 阅读：285 来源： 互联网

标签：Paths PAT heap 1155 tree int Heap path heapJudge

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:

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+="\n";
        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;
}

标签：Paths,PAT,heap,1155,tree,int,Heap,path,heapJudge
来源： https://www.cnblogs.com/littlepage/p/12207175.html

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