编写一个C++ 战棋A星寻路算法

好的，我了解了你的要求，并且明白你的问题。以下是关于编写C语言中的A星寻路算法的基本介绍：

A星算法是一种基于权重的最短路径优先算法，用于寻找从起点到目标节点的最短路径。该算法将节点排列成一个网格图，并采用估价函数（曼哈顿距离或欧几里得距离）来计算每个节点到目标节点的距离。然后，从起点开始，计算每个相邻节点的总成本，并选择具有最低成本的节点进行遍历。最后，重复该过程，直到到达目标节点。

为了编写C语言中的A星寻路算法，需要完成以下步骤：

1. 设计程序接口和数据结构，如节点、地图和路径。

2. 编写算法函数，计算每个节点到目标节点的距离，并选择最佳路径。

3. 编写程序主函数，实现对输入数据的读取和对算法函数的调用。

以下是一个示例程序的代码，该程序实现了A星寻路算法：

#include<stdio.h>
#include<stdlib.h>
#define MAX 100
#define WALL 1
#define EMPTY 0
#define PATH 2
#define GOAL 3
#define START 4
#define M 10
#define N 10

typedef struct _NODE {
    int x, y;
    int f, g, h;
    struct _NODE *parent;
} NODE;

int maze[M][N] = {
    { 0, 0, 1, 0, 0, 0, 1, 0, 0, 0 },
    { 0, 0, 1, 0, 0, 0, 1, 0, 0, 0 },
    { 0, 0, 0, 0, 1, 1, 0, 0, 0, 0 },
    { 0, 0, 0, 0, 1, 0, 0, 0, 0, 0 },
    { 0, 0, 0, 0, 0, 0, 0, 1, 1, 0 },
    { 0, 0, 0, 0, 1, 1, 0, 1, 1, 0 },
    { 0, 0, 0, 0, 0, 0, 0, 1, 0, 0 },
    { 0, 0, 1, 1, 1, 1, 1, 0, 0, 0 },
    { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 },
    { 0, 0, 0, 0, 0, 0, 0, 0, 0, 0 }
};

NODE *open_list[MAX], *closed_list[MAX];
NODE start = {0, 0, 0, 0, 0, NULL};
NODE goal = {9, 9, 0, 0, 0, NULL};
int open_count = 0;
int closed_count = 0;

NODE *find_node_on_list(NODE **list, NODE *node)
{
    int i;
    for(i = 0; list[i] != NULL; i++)
        if(list[i]->x == node->x && list[i]->y == node->y) return list[i];
    return NULL;
}

void remove_node_from_list(NODE **list, NODE *node)
{
    int i;
    for(i = 0; list[i] != NULL; i++) {
        if(list[i] == node) {
            list[i] = NULL;
            return;
        }
    }
}

int compare_nodes(NODE *node1, NODE *node2) {
    if(node1->f < node2->f) return 1; //node1的f小，说明node1的总成本更小
    if(node1->f > node2->f) return -1; //node2的f小，说明node2的总成本更小
    return 0;
}

void insert_node_to_list(NODE **list, NODE *node)
{
    int i;
    for(i = 0; list[i] != NULL && compare_nodes(node, list[i]) > 0; i++);
    for(int j = MAX-2; j-i >= 0; j--) list[j+1] = list[j];
    list[i] = node;
}

int manhattan_distance(NODE *node1, NODE *node2)
{
    return abs(node1->x - node2->x) + abs(node1->y - node2->y);
}

void set_start_and_goal(int sx, int sy, int gx, int gy)
{
    start.x = sx; start.y = sy;
    goal.x = gx; goal.y = gy;
}

NODE *find_path()
{
    NODE *current;

    start.g = 0;
    start.h = manhattan_distance(&start, &goal);
    start.f = start.g + start.h;

    insert_node_to_list(open_list, &start);
    while(open_list[0] != NULL) {
        current = open_list[0];
        if(current->x == goal.x && current->y == goal.y) return current;
        remove_node_from_list(open_list, current);
        insert_node_to_list(closed_list, current);

        for(int i = -1; i <= 1; i++) {
            for(int j = -1; j <= 1; j++) {
                if(i == 0 && j == 0) continue;
                if(current->x+i < 0 || current->x+i >= M || current->y+j < 0 || current->y+j >= N) continue;
                if(maze[current->x+i][current->y+j] == WALL) continue;

                NODE *n = (NODE*)malloc(sizeof(NODE));
                n->x = current->x + i;
                n->y = current->y + j;
                n->parent = current;
                n->g = current->g + 1;
                n->h = manhattan_distance(n, &goal);
                n->f = n->g + n->h;
                if(find_node_on_list(closed_list, n)) continue;
                NODE *opened = find_node_on_list(open_list, n);
                if(!opened) insert_node_to_list(open_list, n);
                else if(n->g < opened->g) {
                    opened->g = n->g;
                    opened->parent = current;
                }
            }
        }
    }
    return NULL;
}

void print_path(NODE *node) {
    if(node == NULL) {
        printf("Cannot find path.\n");
        return;
    }

    int path_length = 0;
    while(node->parent != NULL) {
        maze[node->x][node->y] = PATH;
        node = node->parent;
        path_length++;
    }
    maze[start.x][start.y] = START;
    maze[goal.x][goal.y] = GOAL;

    for (int i = 0; i < M; i++) {
        for (int j = 0; j < N; j++) {
            if (maze[i][j] == WALL) {
                printf("# ");
            } else if (maze[i][j] == EMPTY) {
                printf(". ");
            } else if (maze[i][j] == PATH) {
                printf("* ");
            } else if (maze[i][j] == START) {
                printf("S ");
            } else if (maze[i][j] == GOAL){
                printf("G ");
            }
        }
        printf("\n");
    }
    printf("\nPath length: %d", path_length);
}

int main()
{
    set_start_and_goal(0, 0, 9, 9);

    NODE *goal_node = find_path();

    print_path(goal_node);
    return 0;
}

该程序能够自动解决在一个10*10的网格中的起点和终点之间的最短路径问题。

以上就是这个问题的回答。希望可以帮助到你！