1155 Heap Paths DFS+stack

该代码实现了一棵完全二叉树的构造,并通过深度优先搜索(DFS)进行遍历。同时,它判断了树是否为最大堆或最小堆。输入整数n和节点值,程序会输出树的类型:最大堆、最小堆或非堆。

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代码逻辑很简单
构造一颗完全二叉树 再DFS

#include <cstdio>
#include <stack>
#include <algorithm>
#include <vector>
#include <map>
#include <iostream>
#include <cmath>
using namespace std;
int n;
struct Node
{
    int value;
    Node *lchild, *rchild;
    Node()
    {
        lchild = rchild = NULL;
    }
};
Node N[1010];
int type = 0;

void show(vector<Node> vs)
{
    if (!vs.empty())
    {
        for (int i = 0; i < vs.size(); i++)
        {
            printf("%d", vs[i].value);
            if (i != vs.size() - 1)
            {
                printf(" ");
            }
        }
        printf("\n");
    }
}
vector<Node> s;
void DFS(int index)
{
    if (index > n)
    {
        if (index % 2 == 0)//左孩子都没了 才输出哦
        {
            show(s);
        }
        return;
    }
    s.push_back(N[index]);
    DFS(index * 2 + 1);
    DFS(index * 2);
    s.pop_back();
}
void isMax()
{
    for (int i = 1; i <= n; i++)
    {
        if (i * 2 < n)
        {
            if (N[i * 2].value > N[i].value || N[i * 2 + 1].value > N[i].value)
            {
                type = -1;
                break;
            }
        }
        if (i == n)
        {
            type = 1;
        }
    }
}
void isMin()
{
    for (int i = 1; i <= n; i++)
    {
        if (i * 2 < n)
        {
            if (N[i * 2].value < N[i].value || N[i * 2 + 1].value < N[i].value)
            {
                type = -1;
                break;
            }
        }
        if (i == n)
        {
            type = 2;
        }
    }
}
int main()
{
    scanf("%d", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%d", &N[i].value);
        if (i * 2 <= n)
        {
            N[i].lchild = &N[i * 2];
        }
        if (i * 2 + 1 <= n)
        {
            N[i].rchild = &N[i * 2 + 1];
        }
    }
    isMax();
    if (type == -1)
    {
        isMin();
    }
    if (type == -1)
    {
        type = 0;
    }
    DFS(1);
    if (type == 1)
    {
        printf("Max Heap\n");
    }
    else if (type == 2)
    {
        printf("Min Heap\n");
    }
    else
    {
        printf("Not Heap\n");
    }

    return 0;
}
好的,我会根据您的要求进行回答。首先是邻接矩阵和邻接表的存储结构: 邻接矩阵: ``` 0 1 2 3 4 5 0 0 1 0 0 0 1 1 0 0 1 0 0 0 2 1 0 0 1 0 0 3 0 0 0 0 1 0 4 0 0 1 0 0 1 5 0 0 0 1 1 0 ``` 邻接表: ``` 0 -> 1 -> 5 1 -> 2 2 -> 0 -> 3 3 -> 4 4 -> 2 -> 5 5 -> 3 -> 4 ``` 接下来是深度优先遍历和广度优先遍历算法的实现,以及从顶点5到顶点2的所有长度为3的简单路径: 深度优先遍历算法: ```python def dfs(graph, start): visited = set() stack = [start] while stack: vertex = stack.pop() if vertex not in visited: visited.add(vertex) stack.extend(graph[vertex] - visited) return visited ``` 广度优先遍历算法: ```python from collections import deque def bfs(graph, start): visited = set() queue = deque([start]) while queue: vertex = queue.popleft() if vertex not in visited: visited.add(vertex) queue.extend(graph[vertex] - visited) return visited ``` 从顶点5到顶点2的所有长度为3的简单路径: ```python def find_all_paths(graph, start, end, path=[]): path = path + [start] if start == end: return [path] if start not in graph: return [] paths = [] for node in graph[start]: if node not in path: new_paths = find_all_paths(graph, node, end, path) for p in new_paths: if len(p) == 4 and p[-1] == end: paths.append(p) return paths ``` 使用 `find_all_paths(graph, 5, 2)` 即可输出所有长度为3的简单路径。 最后是从顶点5到顶点2的最短路径,可以使用 Dijkstra 算法或者 Floyd 算法,这里给出 Dijkstra 算法的实现: Dijkstra 算法: ```python import heapq def dijkstra(graph, start, end): heap = [(0, start)] visited = set() while heap: (cost, vertex) = heapq.heappop(heap) if vertex not in visited: visited.add(vertex) if vertex == end: return cost for next_vertex, c in graph[vertex].items(): if next_vertex not in visited: heapq.heappush(heap, (cost + c, next_vertex)) return float("inf") ``` 使用 `dijkstra(graph, 5, 2)` 即可输出从顶点5到顶点2的最短路径长度。 最后是邻接表的销毁,只需将所有链表节点释放即可,示例代码如下: ```python def destroy_graph(graph): for node in graph: graph[node] = None del graph ```
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