8 位数码难题的问题求解

#include<iostream>#include<string>#include<string.h>#include<queue>#include<set>#include<algorithm>using namespace std;struct node{int x,y;int step;//步数char pre;//前一步操作U`D`L`R  ps:上一步操作是U，那么下一步操作不能是D，因为会产生死循环..char map[5][5]; //九宫地图char path[10000];//记录路径};//可理解为当前状态地图信息-结构体char begin[5][5],over[5][5];//开始地图和结束地图 已知.int x1,y1;//记录空格坐标void swap(int *a,int *b){int *t;*t=*a;*a=*b;*b=*t;}bool check(node a){int i,j;for(i=0;i<3;i++)for(j=0;j<3;j++){if(a.map[i][j]!=over[i][j])return false;}return true;}set<string> hsh;bool getvis(node a){string tmp="";int i,j;for(i=0;i<3;i++)for(j=0;j<3;j++){tmp+=a.map[i][j];}if(hsh.find(tmp)!=hsh.end())//如果找到有相同的排序，那么不可进入字符串数组hsh 隐含地把用更多步数而得到相同排序的情况排除！return false;hsh.insert(tmp);return true;}void pfMapPath(node tmp){int i,j,k;cout<<"初始状态"<<endl;for(i=0;i<3;i++){for(j=0;j<3;j++){if(begin[i][j]=='0'){tmp.map[i][j]=' ';cout<<"  ";}else{tmp.map[i][j]=begin[i][j];cout<<begin[i][j]<<" ";}}cout<<endl;}cout<<endl;for(i=1;i<=tmp.step;i++){if(tmp.path[i]=='U'){cout<<"前一状态走向:U"<<" 第"<<i<<"个状态:"<<endl;swap(tmp.map[x1][y1],tmp.map[x1-1][y1]);x1--;for(j=0;j<3;j++){for(k=0;k<3;k++){if(tmp.map[j][k]=='0'){cout<<"  ";}else{cout<<tmp.map[j][k]<<" ";}}cout<<endl;}}else if(tmp.path[i]=='D'){cout<<"前一状态走向:D"<<" 第"<<i<<"个状态:"<<endl;swap(tmp.map[x1][y1],tmp.map[x1+1][y1]);x1++;for(j=0;j<3;j++){for(k=0;k<3;k++){if(tmp.map[j][k]=='0'){cout<<"  ";}else{cout<<tmp.map[j][k]<<" ";}}cout<<endl;}}else if(tmp.path[i]=='R'){cout<<"前一状态走向:R"<<" 第"<<i<<"个状态:"<<endl;swap(tmp.map[x1][y1],tmp.map[x1][y1+1]);y1++;for(j=0;j<3;j++){for(k=0;k<3;k++){if(tmp.map[j][k]=='0'){cout<<"  ";}else{cout<<tmp.map[j][k]<<" ";}}cout<<endl;}}else if(tmp.path[i]=='L'){cout<<"前一状态走向:L"<<" 第"<<i<<"个状态:"<<endl;swap(tmp.map[x1][y1],tmp.map[x1][y1-1]);y1--;for(j=0;j<3;j++){for(k=0;k<3;k++){if(tmp.map[j][k]=='0'){cout<<"  ";}else{cout<<tmp.map[j][k]<<" ";}}cout<<endl;}}}cout<<"终止状态:"<<endl;for(i=0;i<3;i++){for(j=0;j<3;j++){if(over[i][j]=='0'){cout<<"  ";}else{cout<<over[i][j]<<" ";}}cout<<endl;}}bool bfs(int x,int y){node start;start.x=x;start.y=y;start.step=0;start.pre='X';int i,j;for(i=0;i<3;i++)for(j=0;j<3;j++)start.map[i][j]=begin[i][j];queue<node> q;q.push(start);while(!q.empty()){node tmp=q.front(),tmp1;q.pop();if(check(tmp)){cout<<tmp.step<<endl;return true;}tmp1=tmp;if(tmp1.pre!='U'&&tmp1.x+1<3){tmp1.path[tmp1.step]=tmp1.pre;tmp1.pre='D';tmp1.step++;swap(tmp1.map[tmp1.x][tmp1.y],tmp1.map[tmp1.x+1][tmp1.y]);tmp1.x++;if(getvis(tmp1)){if(check(tmp1)){tmp1.path[tmp1.step]=tmp1.pre;pfMapPath(tmp1);//cout<<tmp1.step<<endl;return true;}else{q.push(tmp1);}}}tmp1=tmp;if(tmp1.pre!='D'&&tmp1.x-1>=0){tmp1.path[tmp1.step]=tmp1.pre;tmp1.pre='U';tmp1.step++;swap(tmp1.map[tmp1.x][tmp1.y],tmp1.map[tmp1.x-1][tmp1.y]);tmp1.x--;if(getvis(tmp1)){if(check(tmp1)){tmp1.path[tmp1.step]=tmp1.pre;pfMapPath(tmp1);//cout<<tmp1.step<<endl;return true;}else{q.push(tmp1);}}}tmp1=tmp;if(tmp1.pre!='L'&&tmp1.y+1<3){tmp1.path[tmp1.step]=tmp1.pre;tmp1.pre='R';tmp1.step++;swap(tmp1.map[tmp1.x][tmp1.y],tmp1.map[tmp1.x][tmp1.y+1]);tmp1.y++;if(getvis(tmp1)){if(check(tmp1)){tmp1.path[tmp1.step]=tmp1.pre;pfMapPath(tmp1);//cout<<tmp1.step<<endl;return true;}else{q.push(tmp1);}}}tmp1=tmp;if(tmp1.pre!='R'&&tmp1.y-1>=0){tmp1.path[tmp1.step]=tmp1.pre;tmp1.pre='L';tmp1.step++;swap(tmp1.map[tmp1.x][tmp1.y],tmp1.map[tmp1.x][tmp1.y-1]);tmp1.y--;if(getvis(tmp1)){if(check(tmp1)){tmp1.path[tmp1.step]=tmp1.pre;pfMapPath(tmp1);//cout<<tmp1.step<<endl;return true;}else{q.push(tmp1);}}}}return false;}int main(){char str[100];cin>>str;int x,y;//记录空格坐标int i,j;for(i=0;i<3;i++){for(j=0;j<3;j++){if(str[i*3+j]=='.'){x=i;y=j;x1=x;y1=y;begin[i][j]='0';}elsebegin[i][j]=str[i*3+j];}}cin>>str;for(i=0;i<3;i++){for(j=0;j<3;j++){if(str[i*3+j]=='.')over[i][j]='0';elseover[i][j]=str[i*3+j];}}if(!bfs(x,y))cout<<"-1"<<endl;return 0;}