有界深度优先搜索算法解决八数码问题

《人工智能导论》（第四版）
课后题5.3 C++实现
题目：

运行结果：

C++代码：
思想：Octal结构体表示一个状态，其中parent和current为父状态和当前状态的值，这个值可以理解为ID，唯一标识一个状态，生成的规则在to_number(int [4][4])函数中。变换空格位置，按顺序分别为：上、下、左、右。伪代码为书中P116的伪代码。

#include <iostream>
#include <vector>
using namespace std;struct Octal
{int arr[4][4]; // 存储八数码int parent; // 父亲的值int current; // 当前的值int depth; // 当前深度
};int to_number(int [4][4]);
bool up(int [4][4]);
bool down(int [4][4]);
bool left(int [4][4]);
bool right(int [4][4]);int main()
{int arr[4][4] = {0}; // 存储当前八数码vector<Octal> open; // open表vector<Octal> closed; // closed表int depth; // 深度Octal octal; // 八数码cout << "请输入八数码数据，按照从左到右，从上至下的顺序，空格用0代替：" << endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cin >> arr[i][j];octal.arr[i][j] = arr[i][j];}}cout << "请输入深度：" << endl;cin >> depth;cout << "===开始运算===" << endl;octal.parent = 0;octal.current = to_number(arr);octal.depth = 0;open.push_back(octal);int goal = 123804765; // 最终的八数码值bool flag = false; // 成功解决为true，失败为falseOctal temp;int count = 1;while (!open.empty()){temp.current = open[open.size()-1].current;temp.parent = open[open.size()-1].parent;temp.depth = open[open.size()-1].depth;for (int i = 1; i <= 3; i++)for (int j = 1; j <= 3; j++)temp.arr[i][j] = open[open.size()-1].arr[i][j];open.pop_back();closed.push_back(temp);if (temp.current == goal){flag = true;break;}if (temp.depth >= depth){continue;}Octal produce[4]; // 四个可能生成的八数码数组for (int i = 0; i < 4; i++){produce[i].parent = temp.current;for (int j = 1; j <= 3; j++)for (int k = 1; k <= 3; k++)produce[i].arr[j][k] = temp.arr[j][k];produce[i].depth = temp.depth + 1;}bool flagProduce[4][2]; // 四个八数码对应的标志，0为open中是否含有，1位closed中是否含有for (int i = 0; i < 4; i++)for (int j = 0; j < 2; j++)flagProduce[i][j] = true;if (up(produce[0].arr)){produce[0].current = to_number(produce[0].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[0].current){flagProduce[0][0] = false;break;}if (flagProduce[0][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[0].current){flagProduce[0][1] = false;break;}if (flagProduce[0][1])open.push_back(produce[0]);}/*if (flagProduce[0][0] && flagProduce[0][1]){cout << "up " << count++ << " " << produce[0].depth << endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cout << produce[0].arr[i][j] << " ";}cout << endl;}cout << produce[0].current << endl;cout << "============" << endl;}*/}if (down(produce[1].arr)){produce[1].current = to_number(produce[1].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[1].current){flagProduce[1][0] = false;break;}if (flagProduce[1][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[1].current){flagProduce[1][1] = false;break;}if (flagProduce[1][1])open.push_back(produce[1]);}/*if (flagProduce[1][0] && flagProduce[1][1]){cout << "down " << count++ << " " << produce[1].depth << endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cout << produce[1].arr[i][j] << " ";}cout << endl;}cout << produce[1].current << endl;cout << "============" << endl;}*/}if (left(produce[2].arr)){produce[2].current = to_number(produce[2].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[2].current){flagProduce[2][0] = false;break;}if (flagProduce[2][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[2].current){flagProduce[2][1] = false;break;}if (flagProduce[2][1])open.push_back(produce[2]);}/*if (flagProduce[2][0] && flagProduce[2][1]){cout << "left " << count++ << " " << produce[2].depth << endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cout << produce[2].arr[i][j] << " ";}cout << endl;}cout << produce[2].current << endl;cout << "============" << endl;}*/}if (right(produce[3].arr)){produce[3].current = to_number(produce[3].arr);for (int i = 0; i < open.size(); i++)if (open[i].current == produce[3].current){flagProduce[3][0] = false;break;}if (flagProduce[3][0]){for (int i = 0; i < closed.size(); i++)if (closed[i].current == produce[3].current){flagProduce[3][1] = false;break;}if (flagProduce[3][1])open.push_back(produce[3]);}/*if (flagProduce[3][0] && flagProduce[3][1]){cout << "right " << count++ << " " << produce[3].depth << endl;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){cout << produce[3].arr[i][j] << " ";}cout << endl;}cout << produce[3].current << endl;cout << "============" << endl;}*/}}if (flag){vector<Octal> answer;cout << "找到一个解！" << endl;answer.push_back(temp);int parent = temp.parent;while(parent != 0){for (int j = 0; j < closed.size(); j++){if (closed[j].current == parent){temp.current = closed[j].current;temp.depth = closed[j].depth;temp.parent = closed[j].parent;for (int k = 1; k <= 3; k++){for (int m = 1; m <= 3; m++){temp.arr[k][m] = closed[j].arr[k][m];}}answer.push_back(temp);parent = closed[j].parent;break;}}}for (int i = answer.size()-1; i >= 0; i--){cout << "第" << answer.size()-i-1 << "次：" << endl;for (int j = 1; j <= 3; j++){for (int k = 1; k <= 3; k++){cout << answer[i].arr[j][k] << " ";}cout << endl;}}}else{cout << "在深度限制内的所有结果搜索完毕，没有答案！" << endl;}return 0;
}int to_number(int arr[4][4])
{// 将八数码矩阵中的数据转换为数字，方便存储和比较// 从左到右，从上至下，按照从1到9递增，1为亿位，……，9为个位int index = 100000000;long long result = 0;for (int i = 1; i <= 3; i++){for (int j = 1; j <= 3; j++){result += arr[i][j] * index;index /= 10;}}return result;
}bool up(int arr[4][4])
{// 空格向上int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (i == 1)return false; // 此时无法向上else{arr[i][j] = arr[i-1][j];arr[i-1][j] = 0;}return true;
}bool down(int arr[4][4])
{// 空格向下int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (i == 3)return false; // 此时无法向下else{arr[i][j] = arr[i+1][j];arr[i+1][j] = 0;}return true;
}bool left(int arr[4][4])
{// 空格向左int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (j == 1)return false; // 此时无法向左else{arr[i][j] = arr[i][j-1];arr[i][j-1] = 0;}return true;
}bool right(int arr[4][4])
{// 空格向右int i, j; // 空格的位置为(i, j)bool flag = false;for (i = 1; i <= 3; i++){for (j = 1; j <= 3; j++)if (arr[i][j] == 0){flag = true;break;}if (flag)break;}if (j == 3)return false; // 此时无法向右else{arr[i][j] = arr[i][j+1];arr[i][j+1] = 0;}return true;
}

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