通过二分法用c语言解决八数码问题

八数码问题是指在一个3x3的棋盘上，摆放着编号为1~8的8个方块，再加上一个空格，共计9个空格，如下图所示：

+---+---+---+
| 1 | 2 | 3 |
+---+---+---+
| 4 |   | 5 |
+---+---+---+
| 6 | 7 | 8 |
+---+---+---+

现在需要将这些方块按照规定的顺序移动，使得最终的棋盘状态变为：

+---+---+---+
| 1 | 2 | 3 |
+---+---+---+
| 4 | 5 | 6 |
+---+---+---+
| 7 | 8 |   |
+---+---+---+

其中，每次只能移动与空格相邻的方块，且空格不能移动。现在我们需要用C语言实现通过二分法解决八数码问题。

首先，我们需要定义一个结构体用来表示每一个棋盘状态，其中包括：

- 一个3x3的二维数组，表示当前的棋盘状态；
- 一个整数，表示当前状态到达目标状态的代价估计值（可以用曼哈顿距离或者错误位置数量等方式计算）；
- 一个指向父状态的指针，用来记录状态转移路径。

代码如下：

typedef struct node {
    int board[3][3];
    int cost;
    struct node *parent;
} Node;

接下来，我们需要实现状态转移函数，即根据当前的状态，生成所有可能的下一步状态。这里我们可以定义一个函数`generate_next_states`，它的输入是当前状态的指针，输出是所有可能的下一步状态的指针数组。该函数的实现如下：

Node** generate_next_states(Node *current_state, int *num_states) {
    int x, y, i, j, temp;
    Node **next_states = (Node**)malloc(sizeof(Node*) * 4);
    *num_states = 0;

    // 找到空格的位置
    for (i = 0; i < 3; i++) {
        for (j = 0; j < 3; j++) {
            if (current_state->board[i][j] == 0) {
                x = i;
                y = j;
                break;
            }
        }
    }

    // 生成所有可能的下一步状态
    if (x > 0) { // 上移
        next_states[*num_states] = (Node*)malloc(sizeof(Node));
        memcpy(next_states[*num_states]->board, current_state->board, sizeof(int) * 9);
        temp = next_states[*num_states]->board[x][y];
        next_states[*num_states]->board[x][y] = next_states[*num_states]->board[x-1][y];
        next_states[*num_states]->board[x-1][y] = temp;
        next_states[*num_states]->cost = estimate_cost(next_states[*num_states]->board);
        next_states[*num_states]->parent = current_state;
        (*num_states)++;
    }
    if (x < 2) { // 下移
        next_states[*num_states] = (Node*)malloc(sizeof(Node));
        memcpy(next_states[*num_states]->board, current_state->board, sizeof(int) * 9);
        temp = next_states[*num_states]->board[x][y];
        next_states[*num_states]->board[x][y] = next_states[*num_states]->board[x+1][y];
        next_states[*num_states]->board[x+1][y] = temp;
        next_states[*num_states]->cost = estimate_cost(next_states[*num_states]->board);
        next_states[*num_states]->parent = current_state;
        (*num_states)++;
    }
    if (y > 0) { // 左移
        next_states[*num_states] = (Node*)malloc(sizeof(Node));
        memcpy(next_states[*num_states]->board, current_state->board, sizeof(int) * 9);
        temp = next_states[*num_states]->board[x][y];
        next_states[*num_states]->board[x][y] = next_states[*num_states]->board[x][y-1];
        next_states[*num_states]->board[x][y-1] = temp;
        next_states[*num_states]->cost = estimate_cost(next_states[*num_states]->board);
        next_states[*num_states]->parent = current_state;
        (*num_states)++;
    }
    if (y < 2) { // 右移
        next_states[*num_states] = (Node*)malloc(sizeof(Node));
        memcpy(next_states[*num_states]->board, current_state->board, sizeof(int) * 9);
        temp = next_states[*num_states]->board[x][y];
        next_states[*num_states]->board[x][y] = next_states[*num_states]->board[x][y+1];
        next_states[*num_states]->board[x][y+1] = temp;
        next_states[*num_states]->cost = estimate_cost(next_states[*num_states]->board);
        next_states[*num_states]->parent = current_state;
        (*num_states)++;
    }

    return next_states;
}

其中，`estimate_cost`是一个估价函数，用于估计当前状态到目标状态的代价。这里我们采用曼哈顿距离作为估价函数，其计算方式如下：

int estimate_cost(int board[3][3]) {
    int i, j, x, y, cost = 0;
    int target_x[9] = {0, 0, 0, 1, 1, 1, 2, 2, 2};
    int target_y[9] = {0, 1, 2, 0, 1, 2, 0, 1, 2};

    for (i = 0; i < 3; i++) {
        for (j = 0; j < 3; j++) {
            if (board[i][j] == 0) continue;
            x = target_x[board[i][j]-1];
            y = target_y[board[i][j]-1];
            cost += abs(i - x) + abs(j - y);
        }
    }

    return cost;
}

最后，我们可以实现一个二分法函数`solve_puzzle`，它的输入是初始状态和目标状态，输出是最终的状态转移路径。该函数的实现如下：

Node* solve_puzzle(Node *initial_state, Node *target_state) {
    int num_states, i;
    Node *current_state, **next_states, *min_state;
    int l = 0, r = 100, mid;

    // 初始化初始状态
    initial_state->cost = estimate_cost(initial_state->board);
    initial_state->parent = NULL;

    // 二分法搜索
    while (l <= r) {
        mid = (l + r) / 2;

        // 初始化搜索队列
        Queue *queue = create_queue();
        enqueue(queue, initial_state);

        // BFS搜索
        while (!is_queue_empty(queue)) {
            current_state = dequeue(queue);

            // 到达目标状态，返回路径
            if (memcmp(current_state->board, target_state->board, sizeof(int) * 9) == 0) {
                return current_state;
            }

            // 生成所有可能的下一步状态
            next_states = generate_next_states(current_state, &num_states);

            // 将下一步状态加入搜索队列
            for (i = 0; i < num_states; i++) {
                if (next_states[i]->cost + mid <= r) {
                    enqueue(queue, next_states[i]);
                } else {
                    free(next_states[i]);
                }
            }
            free(next_states);
        }

        // 更新搜索区间
        if (current_state->cost <= mid) {
            min_state = current_state;
            r = current_state->cost;
        } else {
            l = current_state->cost;
        }
    }

    return min_state;
}

其中，`Queue`是一个队列结构体，可以用链表实现。`create_queue`、`enqueue`、`dequeue`和`is_queue_empty`分别用来创建队列、入队、出队和判断队列是否为空。这里我们不再赘述，读者可以自行实现。

至此，我们就完成了通过二分法用C语言解决八数码问题的程序。