(中等) HDU 1043 Eight，经典搜索问题。
Problem Description
The 15-puzzle has been around for over 100 years; even if you don't know it by that name, you've seen it. It is constructed with 15 sliding tiles, each with a number from 1 to 15 on it, and all packed into a 4 by 4 frame with one tile missing. Let's call the missing tile 'x'; the object of the puzzle is to arrange the tiles so that they are ordered as:
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 x
where the only legal operation is to exchange 'x' with one of the tiles with which it shares an edge. As an example, the following sequence of moves solves a slightly scrambled puzzle:
1 2 3 4 1 2 3 4 1 2 3 4 1 2 3 4
5 6 7 8 5 6 7 8 5 6 7 8 5 6 7 8
9 x 10 12 9 10 x 12 9 10 11 12 9 10 11 12
13 14 11 15 13 14 11 15 13 14 x 15 13 14 15 x
r-> d-> r->
The letters in the previous row indicate which neighbor of the 'x' tile is swapped with the 'x' tile at each step; legal values are 'r','l','u' and 'd', for right, left, up, and down, respectively.
Not all puzzles can be solved; in 1870, a man named Sam Loyd was famous for distributing an unsolvable version of the puzzle, and
frustrating many people. In fact, all you have to do to make a regular puzzle into an unsolvable one is to swap two tiles (not counting the missing 'x' tile, of course).
In this problem, you will write a program for solving the less well-known 8-puzzle, composed of tiles on a three by three
arrangement.
这前那篇文章也介绍了一些了，那位大神的文章也写的很好了，我就是来水一下。
首先就是我写的康托展开，用来表示每一个状态，其实对于一个状态 2 3 4 1 5 x 7 6 8来说，把x当作0，就是0到8的一个全排了，那么一共就9！种，这样就是第一个＊8！，第二个＊7！。。。这样下去。。。不过每一个乘的数应该是这个数本身的值减去前面比他小的个数，也就是他在剩下的数里面是第几个。
由一个值转化为一个排列也是如此，每一次找第k个还没有用到的数，就是那个数了。因为这里就只有9个数，直接n^2过一遍就好了。。。（对于很多数，可以用数状数组来做，转化为排列的话就是二分加数状数组。）
然后是减枝，减去无解的情况。对于每一次移动，数的逆序对的奇偶性是不会改变的（这里x不算数），然后的话先求逆序数，如果为奇就直接pass掉。
首先是第一次写，
BFS+康托+剪枝，TLE了。
#include<iostream> #include<cstring> using namespace std; const int maxn=400005; const int Enum=46233; const int jie[9]={1,1,2,6,24,120,720,5040,40320}; int que[400005]; int fa[maxn]; int las,fir; int rem[maxn]; char ans1[maxn]; int cou; int hebing(int (*m)[3]) { bool vis[10]={0}; int cou; int ans=0; for(int i=0;i<9;++i) { cou=0; for(int j=0;j<m[i/3][i%3];++j) if(vis[j]) ++cou; vis[m[i/3][i%3]]=1; ans+=(m[i/3][i%3]-cou)*jie[8-i]; } return ans; } void zhankai(int (*m)[3],int x) { bool vis[10]={0}; int cou; int k; for(int i=0;i<3;++i) for(int j=0;j<3;++j) { cou=x/jie[8-3*i-j]+1; // !!! x%=jie[8-3*i-j]; for(k=0;k<9&&cou;++k) if(vis[k]==0) --cou; vis[k-1]=1; // !!! m[i][j]=k-1; // !!! } } void swap(int &a,int &b) { int temp=a; a=b; b=temp; } bool judge(int x,int y) { if(x<0||y<0||x>2||y>2) return 0; return 1; } void showans() { int temp=Enum; char c; cou=0; while(fa[temp]!=-1) { if(rem[temp]==1) c='d'; else if(rem[temp]==2) c='l'; else if(rem[temp]==3) c='u'; else c='r'; ans1[cou++]=c; temp=fa[temp]; } for(int i=cou-1;i>=0;--i) cout<<ans1[i]; cout<<endl; } void solve(int (*m)[3]) { int ni=0; for(int i=0;i<9;++i) for(int j=0;j<i;++j) if(m[i/3][i%3]&&m[j/3][j%3]) if(m[i/3][i%3]<m[j/3][j%3]) ++ni; if(ni%2) { cout<<"unsolvable\n"; return; } las=fir=0; memset(rem,0,sizeof(rem)); bool ok=0; int temp,t1[3][3],temp1; int x0,y0; temp=hebing(m); rem[temp]=-1; que[las++]=temp; fa[temp]=-1; while(las-fir) { temp=que[fir++]; if(temp==Enum) { ok=1; break; } zhankai(t1,temp); for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(t1[i][j]==0) x0=i,y0=j; if(judge(x0-1,y0)) { swap(t1[x0][y0],t1[x0-1][y0]); temp1=hebing(t1); if(rem[temp1]==0) { rem[temp1]=3; fa[temp1]=temp; que[las++]=temp1; } swap(t1[x0][y0],t1[x0-1][y0]); } if(judge(x0+1,y0)) { swap(t1[x0+1][y0],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { rem[temp1]=1; fa[temp1]=temp; que[las++]=temp1; } swap(t1[x0][y0],t1[x0+1][y0]); } if(judge(x0,y0-1)) { swap(t1[x0][y0-1],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { rem[temp1]=2; fa[temp1]=temp; que[las++]=temp1; } swap(t1[x0][y0],t1[x0][y0-1]); } if(judge(x0,y0+1)) { swap(t1[x0][y0+1],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { rem[temp1]=4; fa[temp1]=temp; que[las++]=temp1; } swap(t1[x0][y0],t1[x0][y0+1]); } } if(ok) showans(); else cout<<"unsolvable"<<endl; } int main() { ios::sync_with_stdio(false); int m[3][3]; char c; while(cin>>c) { m[0][0]=c=='x'?0:c-'0'; cin>>c; m[0][1]=c=='x'?0:c-'0'; cin>>c; m[0][2]=c=='x'?0:c-'0'; for(int i=1;i<3;++i) for(int j=0;j<3;++j) { cin>>c; m[i][j]=c=='x'?0:c-'0'; } solve(m); } return 0; }
然后就是
双向BFS+康托+剪枝，就过了。
#include<iostream> #include<cstdio> #include<cstring> using namespace std; const int maxn=400005; const int Enum=46233; const int jie[9]={1,1,2,6,24,120,720,5040,40320}; const int step[4][2]={{1,0},{0,-1},{-1,0},{0,1}}; int Snum,Mnum; int fa[maxn],son[maxn]; int rem[maxn],Mrem; int que1[maxn],que2[maxn]; int las1,las2,fir1,fir2; char ans[maxn]; int hebing(int (*m)[3]) { bool vis[10]={0}; int cou; int ans=0; for(int i=0;i<9;++i) { cou=0; for(int j=0;j<m[i/3][i%3];++j) if(vis[j]) ++cou; vis[m[i/3][i%3]]=1; ans+=(m[i/3][i%3]-cou)*jie[8-i]; } return ans; } void zhankai(int (*m)[3],int x) { bool vis[10]={0}; int cou; int k; for(int i=0;i<3;++i) for(int j=0;j<3;++j) { cou=x/jie[8-3*i-j]+1; // !!! x%=jie[8-3*i-j]; for(k=0;k<9&&cou;++k) if(vis[k]==0) --cou; vis[k-1]=1; // !!! m[i][j]=k-1; // !!! } } void swap(int &a,int &b) { int temp=a; a=b; b=temp; } bool judge(int x,int y) { if(x<0||y<0||x>2||y>2) return 0; return 1; } void showans() { int cou=0; char c; int tnum=Mnum; while(tnum!=Snum) { if(rem[tnum]==1) c='d'; else if(rem[tnum]==2) c='l'; else if(rem[tnum]==3) c='u'; else c='r'; ans[cou++]=c; tnum=fa[tnum]; } for(int i=cou-1;i>=0;--i) cout<<ans[i]; rem[Mnum]=Mrem; tnum=Mnum; while(tnum!=Enum) { rem[tnum]=-rem[tnum]; if(rem[tnum]==1) c='u'; else if(rem[tnum]==2) c='r'; else if(rem[tnum]==3) c='d'; else c='l'; cout<<c; tnum=son[tnum]; } cout<<endl; } bool bfs() { fir1=fir2=las1=las2=0; int temp,t1[3][3],t2[3][3]; int tnum1,tnum2; int x01,y01,x02,y02; que1[las1++]=Snum; que2[las2++]=Enum; rem[Snum]=100; rem[Enum]=-100; son[Enum]=Enum; fa[Snum]=Snum; while(las1-fir1&&las2-fir2) { tnum1=que1[fir1++]; tnum2=que2[fir2++]; zhankai(t1,tnum1); zhankai(t2,tnum2); for(int i=0;i<3;++i) for(int j=0;j<3;++j) { if(t1[i][j]==0) x01=i,y01=j; if(t2[i][j]==0) x02=i,y02=j; } for(int i=0;i<4;++i) { if(judge(x01+step[i][0],y01+step[i][1])==0) continue; swap(t1[x01][y01],t1[x01+step[i][0]][y01+step[i][1]]); temp=hebing(t1); if(rem[temp]==0) { rem[temp]=i+1; fa[temp]=tnum1; que1[las1++]=temp; } else if(rem[temp]<0) { fa[temp]=tnum1; Mrem=rem[temp]; rem[temp]=i+1; Mnum=temp; return 1; } swap(t1[x01][y01],t1[x01+step[i][0]][y01+step[i][1]]); } for(int i=0;i<4;++i) { if(judge(x02+step[i][0],y02+step[i][1])==0) continue; swap(t2[x02][y02],t2[x02+step[i][0]][y02+step[i][1]]); temp=hebing(t2); if(rem[temp]==0) { rem[temp]=-(i+1); son[temp]=tnum2; que2[las2++]=temp; } else if(rem[temp]>0) { son[temp]=tnum2; Mrem=-(i+1); Mnum=temp; return 1; } swap(t2[x02][y02],t2[x02+step[i][0]][y02+step[i][1]]); } } return 0; } void solve(int (*m)[3]) { int ni=0; for(int i=0;i<9;++i) for(int j=0;j<i;++j) if(m[i/3][i%3]&&m[j/3][j%3]) if(m[i/3][i%3]<m[j/3][j%3]) ++ni; if(ni%2) { cout<<"unsolvable\n"; return; } memset(rem,0,sizeof(rem)); Snum=hebing(m); if(Snum==Enum) { cout<<endl; return; } if(bfs()) showans(); else cout<<"unsolvable\n"; } int main() { ios::sync_with_stdio(false); int m[3][3]; char c; while(cin>>c) { m[0][0]=c=='x'?0:c-'0'; cin>>c; m[0][1]=c=='x'?0:c-'0'; cin>>c; m[0][2]=c=='x'?0:c-'0'; for(int i=1;i<3;++i) for(int j=0;j<3;++j) { cin>>c; m[i][j]=c=='x'?0:c-'0'; } solve(m); } return 0; }
A＊（曼哈顿距离）+康托+剪枝。
#include<iostream> #include<cstring> #include<queue> #include<utility> #include<cmath> #include<cstdio> using namespace std; const int maxn=400005; const int Enum=46233; const int jie[9]={1,1,2,6,24,120,720,5040,40320}; struct state { int F,G,num; state() {} state(int a,int b,int c):F(a),G(b),num(c) {} friend bool operator < (state x,state y) { return x.F>y.F; } }; int fa[maxn]; int rem[maxn]; char ans1[maxn]; int cou; int hebing(int (*m)[3]) { bool vis[10]={0}; int cou; int ans=0; for(int i=0;i<9;++i) { cou=0; for(int j=0;j<m[i/3][i%3];++j) if(vis[j]) ++cou; vis[m[i/3][i%3]]=1; ans+=(m[i/3][i%3]-cou)*jie[8-i]; } return ans; } void zhankai(int (*m)[3],int x) { bool vis[10]={0}; int cou; int k; for(int i=0;i<3;++i) for(int j=0;j<3;++j) { cou=x/jie[8-3*i-j]+1; // !!! x%=jie[8-3*i-j]; for(k=0;k<9&&cou;++k) if(vis[k]==0) --cou; vis[k-1]=1; // !!! m[i][j]=k-1; // !!! } } void swap(int &a,int &b) { int temp=a; a=b; b=temp; } bool judge(int x,int y) { if(x<0||y<0||x>2||y>2) return 0; return 1; } void showans() { int temp=Enum; char c; cou=0; while(fa[temp]!=-1) { if(rem[temp]==1) c='d'; else if(rem[temp]==2) c='l'; else if(rem[temp]==3) c='u'; else c='r'; ans1[cou++]=c; temp=fa[temp]; } for(int i=cou-1;i>=0;--i) cout<<ans1[i]; cout<<endl; } int getH(int (*m)[3]) { int H=0; for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(m[i][j]) H+=abs(i-(m[i][j]-1)/3)+abs(j-(m[i][j]-1)%3); else H+=abs(i-2)+abs(j-2); return H; } void solve(int (*m)[3]) { int ni=0; for(int i=0;i<9;++i) for(int j=0;j<i;++j) if(m[i/3][i%3]&&m[j/3][j%3]) if(m[i/3][i%3]<m[j/3][j%3]) ++ni; if(ni%2) { cout<<"unsolvable\n"; return; } priority_queue <state> que; memset(rem,0,sizeof(rem)); bool ok=0; int temp,t1[3][3],temp1; int x0,y0; int H=0,G; state tsta; temp=hebing(m); que.push(state(H,0,temp)); rem[temp]=-1; fa[temp]=-1; while(!que.empty()) { H=0; tsta=que.top(); que.pop(); G=tsta.G; temp=tsta.num; if(temp==Enum) { ok=1; break; } zhankai(t1,temp); for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(t1[i][j]==0) x0=i,y0=j; if(judge(x0-1,y0)) { swap(t1[x0][y0],t1[x0-1][y0]); temp1=hebing(t1); if(rem[temp1]==0) { H=getH(t1); rem[temp1]=3; fa[temp1]=temp; que.push(state(G+1+H,G+1,temp1)); } swap(t1[x0][y0],t1[x0-1][y0]); } if(judge(x0+1,y0)) { swap(t1[x0+1][y0],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { H=getH(t1); rem[temp1]=1; fa[temp1]=temp; que.push(state(G+H+1,G+1,temp1)); } swap(t1[x0][y0],t1[x0+1][y0]); } if(judge(x0,y0-1)) { swap(t1[x0][y0-1],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { H=getH(t1); rem[temp1]=2; fa[temp1]=temp; que.push(state(G+H+1,G+1,temp1)); } swap(t1[x0][y0],t1[x0][y0-1]); } if(judge(x0,y0+1)) { swap(t1[x0][y0+1],t1[x0][y0]); temp1=hebing(t1); if(rem[temp1]==0) { H=getH(t1); rem[temp1]=4; fa[temp1]=temp; que.push(state(G+H+1,G+1,temp1)); } swap(t1[x0][y0],t1[x0][y0+1]); } } if(ok) showans(); else cout<<"unsolvable"<<endl; } int main() { ios::sync_with_stdio(false); int m[3][3]; char c; while(cin>>c) { m[0][0]=c=='x'?0:c-'0'; cin>>c; m[0][1]=c=='x'?0:c-'0'; cin>>c; m[0][2]=c=='x'?0:c-'0'; for(int i=1;i<3;++i) for(int j=0;j<3;++j) { cin>>c; m[i][j]=c=='x'?0:c-'0'; } solve(m); } return 0; }
双向A＊（曼哈顿距离）+康托+剪枝。
#include<iostream> #include<cstdio> #include<cstring> #include<queue> #include<cmath> using namespace std; const int maxn=400005; const int Enum=46233; const int jie[9]={1,1,2,6,24,120,720,5040,40320}; const int step[4][2]={{1,0},{0,-1},{-1,0},{0,1}}; struct state { int F,G,num; state() {} state(int a,int b,int c):F(a),G(b),num(c) {} friend bool operator < (state x,state y) { return x.F>y.F; } }; int Snum,Mnum; int fa[maxn],son[maxn]; int rem[maxn],Mrem; char ans[maxn]; int hebing(int (*m)[3]) { bool vis[10]={0}; int cou; int ans=0; for(int i=0;i<9;++i) { cou=0; for(int j=0;j<m[i/3][i%3];++j) if(vis[j]) ++cou; vis[m[i/3][i%3]]=1; ans+=(m[i/3][i%3]-cou)*jie[8-i]; } return ans; } void zhankai(int (*m)[3],int x) { bool vis[10]={0}; int cou; int k; for(int i=0;i<3;++i) for(int j=0;j<3;++j) { cou=x/jie[8-3*i-j]+1; // !!! x%=jie[8-3*i-j]; for(k=0;k<9&&cou;++k) if(vis[k]==0) --cou; vis[k-1]=1; // !!! m[i][j]=k-1; // !!! } } void swap(int &a,int &b) { int temp=a; a=b; b=temp; } bool judge(int x,int y) { if(x<0||y<0||x>2||y>2) return 0; return 1; } void showans() { int cou=0; char c; int tnum=Mnum; while(tnum!=Snum) { if(rem[tnum]==1) c='d'; else if(rem[tnum]==2) c='l'; else if(rem[tnum]==3) c='u'; else c='r'; ans[cou++]=c; tnum=fa[tnum]; } for(int i=cou-1;i>=0;--i) cout<<ans[i]; rem[Mnum]=Mrem; tnum=Mnum; while(tnum!=Enum) { rem[tnum]=-rem[tnum]; if(rem[tnum]==1) c='u'; else if(rem[tnum]==2) c='r'; else if(rem[tnum]==3) c='d'; else c='l'; cout<<c; tnum=son[tnum]; } cout<<endl; } int getH(int (*m)[3]) { int H=0; for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(m[i][j]) H+=abs(i-(m[i][j]-1)/3)+abs(j-(m[i][j]-1)%3); else H+=abs(i-2)+abs(j-2); return H; } bool bfs() { priority_queue <state> que1,que2; int temp,t1[3][3],t2[3][3]; int tnum1,tnum2; int x01,y01,x02,y02; state tsta1,tsta2; int H,G1,G2; que1.push(state(0,0,Snum)); que2.push(state(0,0,Enum)); rem[Snum]=100; rem[Enum]=-100; son[Enum]=Enum; fa[Snum]=Snum; while(!que1.empty()&&!que2.empty()) { tsta1=que1.top(); tsta2=que2.top(); que1.pop(); que2.pop(); tnum1=tsta1.num; tnum2=tsta2.num; G1=tsta1.G; G2=tsta2.G; zhankai(t1,tnum1); zhankai(t2,tnum2); for(int i=0;i<3;++i) for(int j=0;j<3;++j) { if(t1[i][j]==0) x01=i,y01=j; if(t2[i][j]==0) x02=i,y02=j; } for(int i=0;i<4;++i) { if(judge(x01+step[i][0],y01+step[i][1])==0) continue; swap(t1[x01][y01],t1[x01+step[i][0]][y01+step[i][1]]); temp=hebing(t1); if(rem[temp]==0) { rem[temp]=i+1; fa[temp]=tnum1; H=getH(t1); que1.push(state(H+G1+1,G1+1,temp)); } else if(rem[temp]<0) { fa[temp]=tnum1; Mrem=rem[temp]; rem[temp]=i+1; Mnum=temp; return 1; } swap(t1[x01][y01],t1[x01+step[i][0]][y01+step[i][1]]); } for(int i=0;i<4;++i) { if(judge(x02+step[i][0],y02+step[i][1])==0) continue; swap(t2[x02][y02],t2[x02+step[i][0]][y02+step[i][1]]); temp=hebing(t2); if(rem[temp]==0) { rem[temp]=-(i+1); son[temp]=tnum2; H=getH(t2); que2.push(state(H+G2+1,G2+1,temp)); } else if(rem[temp]>0) { son[temp]=tnum2; Mrem=-(i+1); Mnum=temp; return 1; } swap(t2[x02][y02],t2[x02+step[i][0]][y02+step[i][1]]); } } return 0; } bool JiOujian(int (*m)[3]) { int ni=0; for(int i=0;i<9;++i) for(int j=0;j<i;++j) if(m[i/3][i%3]&&m[j/3][j%3]) if(m[i/3][i%3]<m[j/3][j%3]) ++ni; return ni%2; } void solve(int (*m)[3]) { if(JiOujian(m)) { cout<<"unsolvable\n"; return; } memset(rem,0,sizeof(rem)); Snum=hebing(m); if(Snum==Enum) { cout<<endl; return; } if(bfs()) showans(); else cout<<"unsolvable\n"; } int main() { ios::sync_with_stdio(false); int m[3][3]; char c; while(cin>>c) { m[0][0]=c=='x'?0:c-'0'; cin>>c; m[0][1]=c=='x'?0:c-'0'; cin>>c; m[0][2]=c=='x'?0:c-'0'; for(int i=1;i<3;++i) for(int j=0;j<3;++j) { cin>>c; m[i][j]=c=='x'?0:c-'0'; } solve(m); } return 0; }
IDA＊+剪枝。
#include<iostream> #include<cstdio> #include<cstring> #include<cmath> using namespace std; const int maxn=400005; const int Enum=46233; const int jie[9]={1,1,2,6,24,120,720,5040,40320}; const int Step[4][2]={{1,0},{0,-1},{-1,0},{0,1}}; int ans[maxn]; int Deep; int Node[3][3]; int x01,y01; void swap(int &a,int &b) { int temp=a; a=b; b=temp; } bool judge() { for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(Node[i][j]&&Node[i][j]!=i*3+j+1) return 0; return 1; } bool JiOujian() { int ni=0; for(int i=0;i<9;++i) for(int j=0;j<i;++j) if(Node[i/3][i%3]&&Node[j/3][j%3]) if(Node[i/3][i%3]<Node[j/3][j%3]) ++ni; return ni%2; } int getH() { int reH=0; for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(Node[i][j]) reH+=abs(i-(Node[i][j]-1)/3)+abs(j-(Node[i][j]-1)%3); else reH+=abs(2-i)+abs(2-j); return reH; } void showans(int G) { for(int i=0;i<G;++i) if(ans[i]==0) cout<<'d'; else if(ans[i]==1) cout<<'l'; else if(ans[i]==2) cout<<'u'; else cout<<'r'; cout<<endl; } bool judgeBian(int x,int y) { if(x<0||y<0||x>2||y>2) return 0; return 1; } bool dfs(int G,int lasStep) { int H=getH(); if(H+G>Deep) return 0; if(judge()) { showans(G); return 1; } for(int i=0;i<4;++i) { if(i%2==lasStep%2&&i!=lasStep) continue; if(judgeBian(x01+Step[i][0],y01+Step[i][1])==0) continue; swap(Node[x01][y01],Node[x01+Step[i][0]][y01+Step[i][1]]); x01+=Step[i][0]; y01+=Step[i][1]; ans[G]=i; if(dfs(G+1,i)) return 1; x01-=Step[i][0]; y01-=Step[i][1]; swap(Node[x01][y01],Node[x01+Step[i][0]][y01+Step[i][1]]); } return 0; } void solve() { if(JiOujian()) { cout<<"unsolvable\n"; return; } Deep=1; while(1) { if(dfs(0,-1)) return; ++Deep; } } int main() { ios::sync_with_stdio(false); char c; while(cin>>c) { Node[0][0]=c=='x'?0:c-'0'; cin>>c; Node[0][1]=c=='x'?0:c-'0'; cin>>c; Node[0][2]=c=='x'?0:c-'0'; for(int i=1;i<3;++i) for(int j=0;j<3;++j) { cin>>c; Node[i][j]=c=='x'?0:c-'0'; } for(int i=0;i<3;++i) for(int j=0;j<3;++j) if(Node[i][j]==0) x01=i,y01=j; solve(); } return 0; }
