From 1714844cf4b4a9694cc0bba6377579fbf9d205bb Mon Sep 17 00:00:00 2001 From: po79yr3at Date: Sun, 26 Nov 2023 21:13:45 +0800 Subject: [PATCH] ADD file via upload --- A8code.c | 225 +++++++++++++++++++++++++++++++++++++++++++++++++++++++ 1 file changed, 225 insertions(+) create mode 100644 A8code.c diff --git a/A8code.c b/A8code.c new file mode 100644 index 0000000..1468609 --- /dev/null +++ b/A8code.c @@ -0,0 +1,225 @@ +#include +#include +#include +#include + +#define N 3 // 定义拼图的维度，这是一个3x3的拼图 +#define RANGE33 for(int i = 0; i < 3; i ++) for(int j = 0; j < 3; j ++) +typedef struct Node { + int puzzle[N][N]; // 存储拼图状态的数组 + struct Node* parent; // 指向父节点的指针，用于追踪路径 + int f, g, h; // A*算法中的 f, g, h 值 +} Node; + +// 创建新的拼图节点 +Node* createNode(int puzzle[N][N]) { + Node* newnode = (Node*)malloc(sizeof(Node)); + RANGE33 + newnode->puzzle[i][j] = puzzle[i][j]; + newnode->parent = NULL; + newnode->f = newnode->g = newnode->h = 0; + return newnode; +} + +// 检查两个拼图状态是否相同 +bool isSamePuzzle(int a[N][N], int b[N][N]) { + //相同则返回true,否则返回false + bool sameflag = true; + RANGE33 + if(a[i][j] != b[i][j]) + sameflag = false; + return sameflag; +} + +// 打印拼图状态 +void printPuzzle(int puzzle[N][N]) { + for(int i = 0; i < 3; i ++){ + for(int j = 0; j < 3; j ++) + printf("%d ", puzzle[i][j]); + putchar('\n'); + } + printf("\n"); +} + + +// 启发函数，计算当前状态到目标状态的估计代价 +int heuristic(Node* current, Node* goal) { + int h = 0; + RANGE33 + if(current->puzzle[i][j] != goal->puzzle[i][j]) + h ++; + return h; +} + +// 移动操作，生成新的拼图状态 +Node* move(Node* current, int dir) { + int key_x, key_y;//记录空白块的位置 + // 找到空白块的位置 + RANGE33 + if(current->puzzle[i][j] == 0){ + key_x = i; + key_y = j; + break; + } + int new_x = key_x, new_y = key_y; + // 根据移动方向更新新块的位置，上下左右移动 + switch(dir) + { + case 0: new_y --; break; + case 1: new_y ++; break; + case 2: new_x --; break; + case 3: new_x ++; break; + } + // 检查新位置是否在边界内 + if( new_x < 0 || new_x >= N || new_y < 0 || new_y >= N) + return NULL; + + // 创建新节点，复制当前拼图状态，并交换块的位置 + Node* new_node = createNode(current->puzzle); + new_node->puzzle[key_x][key_y] = current->puzzle[new_x][new_y]; + new_node->puzzle[new_x][new_y] = 0; + return new_node; +} + +// A*算法，寻找最短路径 +Node* AStar(Node* start, Node* goal) { + Node* OPEN[1000]; // 开放列表，用于存储待探索的节点 + Node* CLOSED[1000]; // 关闭列表，用于存储已探索的节点 + int OPEN_SIZE = 0; // 开放列表的大小 + int CLOSED_SIZE = 0; // 关闭列表的大小 + + OPEN[0] = start; // 将起始节点添加到开放列表 + OPEN_SIZE = 1; // 开放列表的大小设置为1 + CLOSED_SIZE = 0; // 关闭列表的大小设置为0 + + while (OPEN_SIZE > 0) {//对open列表进行操作 + int min_f = OPEN[0]->f;//初始化最小的f + int min_index = 0; + // 查找开放列表中具有最小f值的节点 + for(int i = 0; i < OPEN_SIZE; i ++) + if(OPEN[i]->f < min_f){ + min_index = i; + min_f = OPEN[i]->f; + } + + Node* current = OPEN[min_index]; // 获取具有最小f值的节点 + + // 如果当前节点与目标状态匹配，表示找到解 + if(isSamePuzzle(current->puzzle, goal->puzzle)) + return current; + + //开放列表的大小减1，表示从开放列表中移除了一个节点 + OPEN_SIZE --; + + //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。 + //这是为了优化开放列表的结构。 + Node* temp = OPEN[min_index]; + OPEN[min_index] = OPEN[OPEN_SIZE]; + OPEN[OPEN_SIZE] = temp; + + //将当前节点添加到关闭列表，关闭列表大小加1 + CLOSED[CLOSED_SIZE ++] = current; + + int key = 0; + // 查找当前节点中空白块的位置 + for (int i = 0; i < N; i++) { + for (int j = 0; j < N; j++) { + if (current->puzzle[i][j] == 0) { + key = i * N + j; + break; + } + } + } + + // 尝试四个方向的移动操作 + for (int dir = 0; dir < 4; dir++) { + Node* new_node = move(current, dir); + + if (new_node != NULL && !isSamePuzzle(new_node->puzzle, current->puzzle)) { + //得到对应的g、f、h值 + int gNew = current->g + 1; + int hNew = heuristic(new_node, goal); + int fNew = gNew + hNew; + + bool in_OPEN = false; + int open_index = -1; + // 检查新节点是否在开放列表中 + for (int i = 0; i < OPEN_SIZE; i++) { + if (isSamePuzzle(new_node->puzzle, OPEN[i]->puzzle)) { + in_OPEN = true; + open_index = i; + break; + } + } + + bool in_CLOSED = false; + int close_index = -1; + // 检查新节点是否在关闭列表中 + for (int i = 0; i < CLOSED_SIZE; i++) { + if (isSamePuzzle(new_node->puzzle, CLOSED[i]->puzzle)) { + in_CLOSED = true; + close_index = i; + break; + } + } + //若该节点机不在开放列表中也不在关闭列表中 + if (!in_OPEN && !in_CLOSED) { + //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。 + new_node->g = gNew; + new_node->h = hNew; + new_node->f = fNew; + new_node->parent = current; + // 添加新节点new_node到开放列表，开放列表大小加1 + OPEN[OPEN_SIZE ++] = new_node; + } + //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。 + else if (in_OPEN && fNew < OPEN[open_index]->f) { + OPEN[open_index]->f = fNew; + } + } + } + } + return NULL; // 无解 +} + +// 打印解路径 +void printPath(Node* final) { + if (final == NULL) { + return; + } + printPath(final->parent); // 递归打印路径 + for (int i = 0; i < N; i++) { + if (i%3==0){ + printf("-------\n"); + } + for (int j = 0; j < N; j++) { + printf("%d ", final->puzzle[i][j]); + } + + printf("\n"); + + } +} + +int main() { + //int start[N][N] = {{2, 0, 3}, {1, 8, 4}, {7, 6, 5}}; + //int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}}; + + // int start[N][N] = {{2, 8, 3}, {1, 6, 4}, {7, 0, 5}}; + // int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}}; + + int start[N][N] = {{2, 8, 3}, {1, 0, 4}, {7, 6, 5}}; + int target[N][N] = {{1, 2, 3}, {8, 0, 4}, {7, 6, 5}}; + Node* init = createNode(start); + Node* goal = createNode(target); + + Node* final = AStar(init, goal); + if (final) { + printf("This problem has a solution:\n"); + printPath(final); // 打印解路径 + } else { + printf("This problem has no solution!\n"); + } + + return 0; +}