PAT甲级2018春季7-4 1147 Heaps (30分)

清疚 2023-02-24 08:54 25阅读 0赞

[算法笔记总目录][Link 1]  
[关键英语单词解释][Link 2]  
[1147 Heaps (30分)][1147 Heaps _30]  
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]

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：**  
给出一个完全二叉树，判断是不是堆，是什么堆，输出其后序遍历  
**思路：**  
根据完全二叉树的性质 父x 左x*2 右x*2+1 建树，判断堆，直接利用层序数组里 level\[i\] 大于还是小于 level\[2*i\] 与 level\[2*i+1\],在后序遍历

[相似题及思路可参考A1155][A1155]

## 代码一（不建树） ##

#include<bits/stdc++.h>
    using namespace std;
    int n,level[1010],x=0;
    vector<int> v;
    void dfs(int index){ 
        if (index > n ) return;
        v.push_back(level[index]);
        dfs(index * 2 + 1);
        dfs(index * 2);
    }
    int main(){ 
    	scanf("%d %d",&x,&n);
    	for(int j=0;j<x;j++){ 
    		int isMin = 1, isMax = 1;
    		v.clear();
    	    for (int i = 1; i <= n; i++) scanf("%d", &level[i]);
    	    dfs(1);
            for (int i = 2; i <= n; i++) { 
               if (level[i/2] > level[i]) isMin = 0;
               if (level[i/2] < level[i]) isMax = 0;
            }
            if (isMin == 1)  printf("Min Heap");
            else  printf("%s", isMax == 1 ? "Max Heap" : "Not Heap");
            printf("\n");
            printf("%d",v[v.size()-1]);
            for(int i=v.size()-2;i>=0;i--)printf(" %d",v[i]);
            if(j<x-1)printf("\n");
    	}
        return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]

## 代码二（建树） ##

#include<bits/stdc++.h>
    using namespace std;
    struct node{ 
    	int val;
    	node *lchild,*rchild;
    };
    int n,level[1010],x=0,flag=0;
    node *create(int x,node *root)//1号位开始 
    { 
    	if(x>n) return NULL;
    	root->val=level[x];
        root->lchild=create(x*2,new node);
        root->rchild=create(x*2+1,new node);
    	return root;
    }
    
    void printTree(node *root){ 
    	if(root == NULL) return;
    	printTree(root->lchild);
        printTree(root->rchild);
    	if(flag==1)printf(" %d",root->val);
    	else printf("%d",root->val);
    	flag=1;
    }
    int main(){ 
    	scanf("%d %d",&x,&n);
    	for(int j=0;j<x;j++){ 
    		int isMin = 1, isMax = 1;
    		flag=0;
    		node *root = new node;
    	    for (int i = 1; i <= n; i++) scanf("%d", &level[i]);
            for (int i = 2; i <= n; i++) { 
               if (level[i/2] > level[i]) isMin = 0;
               if (level[i/2] < level[i]) isMax = 0;
            }
            if (isMin == 1)  printf("Min Heap");
            else  printf("%s", isMax == 1 ? "Max Heap" : "Not Heap");
            printf("\n");
            printTree(create(1,root));	
            if(j<x-1)printf("\n");
    	}
        return 0;
    }

![在这里插入图片描述][watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70 1]

[Link 1]: https://blog.csdn.net/qq_21480607/article/details/106660085
[Link 2]: https://blog.csdn.net/qq_21480607/article/details/106987591
[1147 Heaps _30]: https://pintia.cn/problem-sets/994805342720868352/problems/994805342821531648
[https_en.wikipedia.org_wiki_Heap_data_structure]: https://en.wikipedia.org/wiki/Heap_%28data_structure
[A1155]: https://blog.csdn.net/qq_21480607/article/details/107219644
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70]: /images/20230209/a21c054ad7ff4c8588825bd25ea50d10.png
[watermark_type_ZmFuZ3poZW5naGVpdGk_shadow_10_text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3_size_16_color_FFFFFF_t_70 1]: https://img-blog.csdnimg.cn/20200709085450732.png?x-oss-process=image/watermark,type_ZmFuZ3poZW5naGVpdGk,shadow_10,text_aHR0cHM6Ly9ibG9nLmNzZG4ubmV0L3FxXzIxNDgwNjA3,size_16,color_FFFFFF,t_70