C. Watto and Mechanism time limit per test3 seconds memory limit per test256 megabytes inputstandard input outputstandard output Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: “Given string s, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs from s in exactly one position”.
Watto has already compiled the mechanism, all that’s left is to write a program for it and check it on the data consisting of n initial lines and m queries. He decided to entrust this job to you.
Input The first line contains two non-negative numbers n and m (0 ≤ n ≤ 3·105, 0 ≤ m ≤ 3·105) — the number of the initial strings and the number of queries, respectively.
Next follow n non-empty strings that are uploaded to the memory of the mechanism.
Next follow m non-empty strings that are the queries to the mechanism.
The total length of lines in the input doesn’t exceed 6·105. Each line consists only of letters ‘a’, ‘b’, ‘c’.
Output For each query print on a single line “YES” (without the quotes), if the memory of the mechanism contains the required string, otherwise print “NO” (without the quotes).
Examples input 2 3 aaaaa acacaca aabaa ccacacc caaac output YES NO NO 字典树+dfs，就当一个字典树的模板吧。。
#include<stdio.h> #include<iostream> #include<algorithm> #include<string> #include<string.h> #include<queue> using namespace std; #define MAXN 1010000 #define F(x,a,b) for (int x=a;x<=b;x++) #define ch(x) ((x)-'a') struct p{int next[10],val;}tre[MAXN]; int cnt,len,n,m; int head[MAXN]; char str[610000]; void insertt() { int tr=1; F(i,0,len-1) { int t=ch(str[i]); if (!tre[tr].next[t]) tre[tr].next[t]=++cnt; tr=tre[tr].next[t]; } tre[tr].val=1; } int anss=0; int _s(int root,int pos,int deep) { int tr;//写成 int tr=root; 后把下面所有root换成tr 不知为何会莫名其妙的出错，求大神解释下。。 if ((deep==1)&&(tre[root].val==1)&&pos==len) return 1; if (pos>len-1) return 0; int t=ch(str[pos]); F(j,0,2) { if(tre[root].next[j]) { tr=tre[root].next[j]; if (ch(str[pos])==j) { if(_s(tr,pos+1,deep)) return 1; } else { if(deep<1&&_s(tr,pos+1,deep+1)) return 1; } } } return 0; } int main() { scanf("%d%d",&n,&m);cnt=1; F(i,1,n){scanf("%s",str);len=strlen(str);insertt();} F(i,1,m){scanf("%s",str);len=strlen(str); if (_s(1,0,0)) printf("YES\n");else printf("NO\n");} }