用盲目搜索技术解决八数码问题

用盲目搜索技术解决八数码问题题目在3×3的棋盘，有八个棋子，每个棋子上标有1至8的某一数字，不同棋子上标的数字不相同。

棋盘上还有一个空格，与空格相邻的棋子可以移到空格中。

要解决的问题是：任意给出一个初始状态和一个目标状态，找出一种从初始转变成目标状态的移动棋子步数最少的移动步骤。

算法流程使用宽度优先搜索从初始节点开始，向下逐层对节点进形依次扩展，并考察它是否为目标节点，再对下层节点进行扩展（或搜索）之前，必须完成对当层的所有节点的扩展。

再搜索过程中，未扩展节点表OPEN中的节点排序准则是：先进入的节点排在前面，后进入的节点排在后面。

宽度优先算法如下：把初始结点S0放入OPEN表中若OPEN表为空，则搜索失败，问题无解取OPEN表中最前面的结点N放在CLOSE表中，并冠以顺序编号n若目标结点Sg N，则搜索成功，问题有解若N无子结点，则转2扩展结点N，将其所有子结点配上指向N的放回指针，依次放入OPEN表的尾部，转2源程序#include <iostream>#include <ctime>#include <vector>using namespace std;const int ROW = 3;//行数const int COL = 3;//列数const int MAXDISTANCE = 10000;//最多可以有的表的数目const int MAXNUM = 10000;typedef struct _Node{int digit[ROW][COL];int dist;//distance between one state and the destination 一个表和目的表的距离int dep; // the depth of node深度// So the comment function = dist + dep.估价函数值int index; // point to the location of parent父节点的位置} Node;Node src, dest;// 父节表目的表vector<Node> node_v; // store the nodes存储节点bool isEmptyOfOPEN() //open表是否为空{for (int i = 0; i < node_v.size(); i++) {if (node_v[i].dist != MAXNUM)return false;}return true;}bool isEqual(int index,int digit[][COL]) //判断这个最优的节点是否和目的节点一样{for (int i = 0; i < ROW; i++)for (int j = 0; j < COL; j++) {if (node_v[index].digit[i][j] != digit[i][j])return false;}return true;}ostream& operator<<(ostream& os, Node& node){for (int i = 0; i < ROW; i++) {for (int j = 0; j < COL; j++)os << node.digit[i][j] << ' ';os << endl;}return os;}void PrintSteps(int index, vector<Node>& rstep_v)//输出每一个遍历的节点深度遍历{rstep_v.push_back(node_v[index]);index = node_v[index].index;while (index != 0){rstep_v.push_back(node_v[index]);index = node_v[index].index;}for (int i=rstep_v.size()-1;i>=0;i--)//输出每一步的探索过程cout << "Step " << rstep_v.size() - i<< endl << rstep_v[i] << endl;}void Swap(int& a, int& b){int t;t = a;a = b;b = t;}void Assign(Node& node, int index){for (int i = 0; i < ROW; i++)for (int j = 0; j < COL; j++)node.digit[i][j] = node_v[index].digit[i][j];}int GetMinNode() //找到最小的节点的位置即最优节点{int dist = MAXNUM;int loc; // the location of minimize nodefor (int i = 0; i < node_v.size(); i++){if (node_v[i].dist == MAXNUM)continue;else if ((node_v[i].dist + node_v[i].dep) < dist) { loc = i;dist = node_v[i].dist + node_v[i].dep;}}return loc;}bool isExpandable(Node& node){for(int i=0;i<node_v.size();i++) {if (isEqual(i, node.digit))return false;}return true;}int Distance(Node& node, int digit[][COL]) {int distance = 0;bool flag = false;for(int i = 0; i < ROW; i++)for (int j = 0; j < COL; j++)for (int k = 0; k < ROW; k++) {for (int l = 0; l < COL; l++) {if (node.digit[i][j] == digit[k][l]) {distance += abs(i - k) + abs(j - l);flag = true;break;}elseflag = false;}if(flag)break;}return distance;}int MinDistance(int a,int b){return (a<b?a:b);}void ProcessNode(int index){int x, y;bool flag;for (int i = 0; i < ROW; i++) {for (int j = 0; j < COL; j++) {if (node_v[index].digit[i][j] == 0) {x =i; y = j;flag = true;break;}else flag = false;}if(flag)break;}Node node_up;Assign(node_up, index);//向上扩展的节点int dist_up = MAXDISTANCE;if (x > 0){Swap(node_up.digit[x][y], node_up.digit[x - 1][y]);if (isExpandable(node_up)){dist_up = Distance(node_up, dest.digit);node_up.index = index;node_up.dist = dist_up;node_up.dep = node_v[index].dep + 1;node_v.push_back(node_up);}}Node node_down;Assign(node_down, index);//向下扩展的节点int dist_down = MAXDISTANCE;if (x < 2){Swap(node_down.digit[x][y], node_down.digit[x + 1][y]); if (isExpandable(node_down)){dist_down = Distance(node_down, dest.digit);node_down.index = index;node_down.dist = dist_down;node_down.dep = node_v[index].dep + 1;node_v.push_back(node_down);}}Node node_left;Assign(node_left, index);//向左扩展的节点int dist_left = MAXDISTANCE;if (y > 0){Swap(node_left.digit[x][y], node_left.digit[x][y - 1]); if (isExpandable(node_left)){dist_left = Distance(node_left, dest.digit);node_left.index = index;node_left.dist = dist_left;node_left.dep = node_v[index].dep + 1;node_v.push_back(node_left);}}Node node_right;Assign(node_right, index);//向右扩展的节点int dist_right = MAXDISTANCE;if (y < 2){Swap(node_right.digit[x][y], node_right.digit[x][y + 1]); if (isExpandable(node_right)){dist_right = Distance(node_right, dest.digit);node_right.index = index;node_right.dist = dist_right;node_right.dep = node_v[index].dep + 1;node_v.push_back(node_right);}}node_v[index].dist = MAXNUM;}int main() // 主函数{int number;cout << "Input source:" << endl;for (int i = 0; i < ROW; i++)//输入初始的表for (int j = 0; j < COL; j++) {cin >> number;src.digit[i][j] = number;}src.index = 0;src.dep = 1;cout << "Input destination:" << endl;//输入目的表for (int m = 0; m < ROW; m++)for (int n = 0; n < COL; n++) {cin >> number;dest.digit[m][n] = number;}node_v.push_back(src);//在容器的尾部加一个数据cout << "Search..." << endl;clock_t start = clock();while (1){if (isEmptyOfOPEN()){cout << "Cann't solve this statement!" << endl;return -1;}else{int loc; // the location of the minimize node最优节点的位置loc = GetMinNode();if(isEqual(loc, dest.digit)){vector<Node> rstep_v;cout << "Source:" << endl;cout << src << endl;PrintSteps(loc, rstep_v);cout << "Successful!" << endl;Cout <<"Using "<<(clock()-start)/CLOCKS_PER_SEC << " seconds." << endl;break;}elseProcessNode(loc);}}return 0;}。