回溯法求解八数码问题

code：

 

#include<cstdio>int path[4];int backtrack(int n){if(n==4){for(int i=0;i<4;i++)printf("%d%d ",i+1,path[i]);printf("/n");return 0;}for(int i=1;i<5;i++){int ok=1;for(int j=0;j<n;j++)if(i==path[j]||i-n==path[j]-j||path[j]-n+j==i){ok=0;break;}if(ok){path[n]=i;backtrack(n+1);}}return 0;}int main(){backtrack(0);return 0;}