18年冬季第四题 PAT甲级 1155 Heap Paths (30分)

太过爱你忘了你带给我的痛 2023-07-07 11:09 32阅读 0赞

# 汇总贴 #

[2020年3月PAT甲级满分必备刷题技巧][2020_3_PAT]

# 题目 #

题目来源：[https://pintia.cn/problem-sets/994805342720868352/problems/1071785408849047552][https_pintia.cn_problem-sets_994805342720868352_problems_1071785408849047552]

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][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<N≤1,000), 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

# 题目分析 #

1.level order traversal sequence of a complete binary tree  
因为输入是层序的完全二叉树，直接存数组里，用DFS就能遍历从根到叶了。

2.all the paths in its right subtree must be printed before those in its left subtree  
先打印右子树，再打印左子树。这个在DFS函数里写好就行。

3. liuchuo的DFS函数跟我的有点不同，稍微复杂些，也可以看看：[https://blog.csdn.net/liuchuo/article/details/84973009][https_blog.csdn.net_liuchuo_article_details_84973009]

# 满分代码 #

#include<iostream>
    #include<vector>
    using namespace std;
    const int maxn=1005;
    vector<int> v(maxn);
    vector<int> vf;
    int n;
    void f(int s){//打印经过第s节点的所有路径
        if(s>n){
            for(int i=0;i<vf.size();i++){
                if(i!=0)printf(" ");
                printf("%d",vf[i]);
            }
            printf("\n");
            return;
        }
        vf.push_back(v[s]);
        if(s*2+1<=n)f(s*2+1);//如果s有右孩子，就去打印经过右孩子的路径
        f(s*2);//如果s有左孩子，同上，如果没有，就直接打印以s为终点的路径
        vf.pop_back();
    }
    int main(){
        int minheap=1,maxheap=1;
        scanf("%d",&n);
        for(int i=1;i<=n;i++){
            scanf("%d",&v[i]);
            if(i>1 && v[i/2]<v[i])maxheap=0;
            if(i>1 && v[i/2]>v[i])minheap=0;
        }
        f(1);
        if(minheap==1)printf("Min Heap\n");
        else printf("%s\n",maxheap==1?"Max Heap":"Not Heap");
        return 0;
    }

[2020_3_PAT]: https://blog.csdn.net/allisonshing/article/details/104213959#comments
[https_pintia.cn_problem-sets_994805342720868352_problems_1071785408849047552]: https://pintia.cn/problem-sets/994805342720868352/problems/1071785408849047552
[https_en.wikipedia.org_wiki_Heap_data_structure]: https://en.wikipedia.org/wiki/Heap_%28data_structure
[https_blog.csdn.net_liuchuo_article_details_84973009]: https://blog.csdn.net/liuchuo/article/details/84973009