uva 129 Kryptopn Factor 暴力回溯

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cur之前无相同相邻子串,path[cur]加入后,只需判断后缀cur是否存在相邻重复子串即可

#include <iostream>#include <cstring>#include <cstdio>#include <algorithm>using namespace std;const int N=1e4+20;char path[N];int cnt;int n,l;bool check(int cur){for(int l=1;l<=(cur+1)/2;l++)// len of the repeted sub{int equal=1;//cur之前无相同相邻子串 for(int i=0;i<l;i++)//判断后缀cur是否存在相邻重复子串即可 {if(path[cur-i]!=path[cur-l-i]){equal=0;break;}}if(equal)return false;}return true;}int dfs(int cur){if(cnt==n){for(int i=0;i<cur;i++){if(i&&i%4==0){if(i%64==0)printf("\n");elseprintf(" ");}printf("%c",path[i]);}printf("\n%d\n",cur);return 1;}for(int i=0;i<l;i++){path[cur]='A'+i;if(check(cur))//add s[i] to path{cnt++;if(dfs(cur+1))return 1;}}return 0;}int main(){while(cin>>n>>l&&(n+l))//n'th {cnt=0;dfs(0);}return 0;}