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#include <stdio.h>
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#include <stdlib.h>
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#include <stdbool.h>
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#include <string.h>
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#define N 3 // 定义拼图的维度，这是一个3x3的拼图
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typedef struct Node {
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int puzzle[N][N]; // 存储拼图状态的数组
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struct Node* parent; // 指向父节点的指针，用于追踪路径
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int f, g, h; // A*算法中的 f, g, h 值
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} Node;
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// 创建新的拼图节点
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Node* createNode(int puzzle[N][N]) {
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Node* newnode = (Node*)malloc(sizeof(Node));
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int i,j;
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for(i=0;i<N;i++)
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{
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for(j=0;j<N;j++)
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{
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newnode->puzzle[i][j]=puzzle[i][j];
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}
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}
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newnode->parent=NULL;
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newnode->f=0;
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newnode->g=0;
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newnode->h=0;
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//请实现该函数
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}
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// 检查两个拼图状态是否相同
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bool isSamePuzzle(int a[N][N], int b[N][N]) {
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int i,j;
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for(i=0;i<N;i++)
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{
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for(j=0;j<N;j++)
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{
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if(a[i][j]!=b[i][j])
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{
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return false;
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}
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}
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}
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return true;
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//相同则返回true,否则返回false
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}
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// 打印拼图状态
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void printPuzzle(int puzzle[N][N]) {
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int i,j;
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for(i=0;i<N;i++)
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{
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for(j=0;j<N;j++)
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{
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printf("%d",puzzle[i][j]);
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if(j<2)
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{
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printf(" ");
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}
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else
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{
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printf("\n");
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}
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}
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}
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//双重for循环实现拼图的打印
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printf("\n");
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}
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// 启发函数，计算当前状态到目标状态的估计代价
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int heuristic(Node* current, Node* goal) {
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int h = 0;
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int i,j;
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for(i=0;i<N;i++)
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{
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for(j=0;j<N;j++)
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{
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if(current->puzzle[i][j]!=goal->puzzle[i][j])
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{
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h++;
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}
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}
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}
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return h;
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// 计算不匹配的拼图块数量
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}
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// 移动操作，生成新的拼图状态555555555555555555555555555555555555555
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Node* move(Node* current, int dir) {
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int key_x, key_y;//记录空白块的位置
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int i,j;// 找到空白块的位置
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for(i=0;i<N;i++)
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{
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for(j=0;j<N;j++)
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{
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if(current->puzzle[i][j]==0)
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{
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key_x=i;
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key_y=j;
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break;
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}
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}
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}
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int new_x,new_y;
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new_x=key_x;
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new_y=key_y;//给new_x、new_y赋值
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if(dir==0)
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{
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new_x--;
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}
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else if(dir==1)
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{
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new_x++;
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}
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else if(dir==2)
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{
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new_y--;
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}
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else if(dir==3)
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{
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new_y++;
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}
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if(new_x<0||new_x>2||new_y<0||new_y>2)
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{
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return NULL;
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}
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// 根据移动方向更新新块的位置，上下左右移动
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// 检查新位置是否在边界内
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// 创建新节点，复制当前拼图状态，并交换块的位置
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Node* new_node = createNode(current->puzzle);
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new_node->puzzle[key_x][key_y] = current->puzzle[new_x][new_y];
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new_node->puzzle[new_x][new_y] = 0;
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return new_node;
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}
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// A*算法，寻找最短路径666666666666666666666666
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Node* AStar(Node* start, Node* goal) {
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Node* OPEN[1000]; // 开放列表，用于存储待探索的节点
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Node* CLOSED[1000]; // 关闭列表，用于存储已探索的节点
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int OPEN_SIZE = 0; // 开放列表的大小
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int CLOSED_SIZE = 0; // 关闭列表的大小
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OPEN[0] = start; // 将起始节点添加到开放列表
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OPEN_SIZE = 1; // 开放列表的大小设置为1
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CLOSED_SIZE = 0; // 关闭列表的大小设置为0
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while (OPEN_SIZE > 0) {//对open列表进行操作
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int min_f = OPEN[0]->f;//初始化最小的f
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int min_index = 0;
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for(int i=0;i<OPEN_SIZE;i++)
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{
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if(OPEN[i]->f<min_f)
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{
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min_f=OPEN[i]->f;
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min_index=i;
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}
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}
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// 查找开放列表中具有最小f值的节点
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Node* current = OPEN[min_index]; // 获取具有最小f值的节点
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if(isSamePuzzle(current->puzzle,goal->puzzle))
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{
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return(current);
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}
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// 如果当前节点与目标状态匹配，表示找到解
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OPEN_SIZE--;
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//开放列表的大小减1，表示从开放列表中移除了一个节点
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//将最小 f 值的节点移到开放列表的末尾，以便稍后将其添加到关闭列表中。
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//这是为了优化开放列表的结构。
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Node* temp = OPEN[min_index];
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OPEN[min_index] = OPEN[OPEN_SIZE];
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OPEN[OPEN_SIZE] = temp;
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//将当前节点添加到关闭列表，关闭列表大小加1
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CLOSED+CLOSED_SIZE=current;
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CLOSED_SIZE++;
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int key = 0;
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// 查找当前节点中空白块的位置
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for (int i = 0; i < N; i++) {
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for (int j = 0; j < N; j++) {
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if (current->puzzle[i][j] == 0) {
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key = i * N + j;
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break;
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}
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}
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}
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// 尝试四个方向的移动操作
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for (int dir = 0; dir < 4; dir++) {
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Node* new_node = move(current, dir);
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if (new_node != NULL && !isSamePuzzle(new_node->puzzle, current->puzzle)) {
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//得到对应的g、f、h值
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int gNew = current->g + 1;
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int hNew = heuristic(new_node, goal);
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int fNew = gNew + hNew;
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bool in_OPEN = false;
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int open_index = -1;
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// 检查新节点是否在开放列表中
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for (int i = 0; i < OPEN_SIZE; i++) {
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if (isSamePuzzle(new_node->puzzle, OPEN[i]->puzzle)) {
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in_OPEN = true;
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open_index = i;
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break;
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}
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}
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bool in_CLOSED = false;
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int closed_index = -1;
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// 检查新节点是否在关闭列表中
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for (int i = 0; i < CLOSED_SIZE; i++) {
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if (isSamePuzzle(new_node->puzzle, CLOSED[i]->puzzle)) {
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in_CLOSED = true;
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closed_index = i;
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break;
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}
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}
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//若该节点机不在开放列表中也不在关闭列表中
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if (!in_OPEN && !in_CLOSED) {
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new_node->g=gNew;
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new_node->h=hNew;
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new_node->f=fNew;
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current=CLOSED+CLOSED_SIZE-1;
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//把gNew、hNew、fNew赋给new_nod对应的g、h、f值，并将其父节点设置为当前节点。
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OPEN[OPEN_SIZE+1]=*new_node;
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OPEN_SIZE++;
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// 添加新节点new_node到开放列表，开放列表大小加1
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}
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//如果新节点已经在开放列表中，但新的 f 值更小，将更新开放列表中已存在节点的信息。
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else if (in_OPEN && fNew < OPEN[open_index]->f) {
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OPEN[open_index]->f=fNew;
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OPEN[open_index]->g=gNew;
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OPEN[open_index]->h=hNew;
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}
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}
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}
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}
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return NULL; // 无解
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}
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// 打印解路径
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void printPath(Node* final) {
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if (final == NULL) {
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return;
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}
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printPath(final->parent); // 递归打印路径
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for (int i = 0; i < N; i++) {
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if (i%3==0){
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printf("-------\n");
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}
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for (int j = 0; j < N; j++) {
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printf("%d ", final->puzzle[i][j]);
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}
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printf("\n");
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}
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}
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int main() {
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//int start[N][N] = {{2, 0, 3}, {1, 8, 4}, {7, 6, 5}};
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//int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}};
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// int start[N][N] = {{2, 8, 3}, {1, 6, 4}, {7, 0, 5}};
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// int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}};
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int start[N][N] = {{2, 8, 3}, {1, 0, 4}, {7, 6, 5}};
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int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}};
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Node* init = createNode(start);
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Node* goal = createNode(target);
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Node* final = AStar(init, goal);
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if (final) {
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printf("This problem has a solution:\n");
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printPath(final); // 打印解路径
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} else {
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printf("This problem has no solution！\n");
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}
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return 0;
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}
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