帮我写一个实现A*算法的c++代码

好的，以下是一个简单的使用C语言实现A*算法的示例代码：

#include <stdio.h>
#include <stdlib.h>
#include <stdbool.h>
#include <math.h>

// 定义地图大小
#define ROWS 10
#define COLS 10

// 定义节点结构体
typedef struct Node {
    int x;          // x坐标
    int y;          // y坐标
    int f;          // f = g + h
    int g;          // 从起点到该点的实际代价
    int h;          // 从该点到终点的估计代价
    struct Node *parent;    // 父节点
} Node;

// 定义地图
int map[ROWS][COLS] = {
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0},
    {0, 0, 0, 0, 0, 0, 0, 0, 0, 0}
};

// 定义起点和终点
Node *start = NULL;
Node *end = NULL;

// 计算两点之间的曼哈顿距离
int manhattan_distance(int x1, int y1, int x2, int y2) {
    return abs(x1 - x2) + abs(y1 - y2);
}

// 获取周围可行走的点
Node **get_neighbors(Node *current) {
    int x = current->x;
    int y = current->y;
    Node **neighbors = malloc(8 * sizeof(Node *));
    int i = 0;
    for (int dx = -1; dx <= 1; dx++) {
        for (int dy = -1; dy <= 1; dy++) {
            if (dx == 0 && dy == 0) {
                continue;
            }
            int nx = x + dx;
            int ny = y + dy;
            if (nx < 0 || nx >= ROWS || ny < 0 || ny >= COLS) {
                continue;
            }
            if (map[nx][ny] == 1) {
                continue;
            }
            Node *neighbor = malloc(sizeof(Node));
            neighbor->x =