题目链接
好久不敲字典树了,都差点忘记了,这道题就是字典树加上DFS搜素即可,dfs搜的时候还要加上条件,就是是否有改变,且最后是否等长,且要刚好改变一个字符。
我用val判断是否是终点,在用nex[]数组往下推。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <algorithm>
#include <limits>
#include <vector>
#include <stack>
#include <queue>
#include <set>
#include <map>
#define lowbit(x) ( x&(-x) )
#define pi 3.141592653589793
#define e 2.718281828459045
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
const int maxN=6e5+5;
int N, M;
char s[maxN];
struct Trie
{
int val; //终点判断
Trie *nex[3];
Trie()
{
val=0;
memset(nex, 0, sizeof(nex));
}
};
Trie *root = new Trie;
void Creat_trie(Trie *rt, int len)
{
if(s[len]!='\0')
{
if(rt->nex[s[len] - 'a'] == 0)
{
rt->nex[s[len] - 'a'] = new Trie;
}
Creat_trie(rt->nex[s[len] - 'a'], len+1);
}
else rt->val=1;
}
bool dfs(Trie *rt, int len, int rest)
{
if(s[len+1] == '\0')
{
if(rest==0 && rt->val==1) return true;
else return false;
}
if(rest == 1)
{
if(rt->nex[s[len+1]-'a']!=0 && dfs(rt->nex[s[len+1]-'a'], len+1, 1)) return true;
for(int i=0; i<3; i++)
{
if(i != s[len+1]-'a' && rt->nex[i]!=0 && dfs(rt->nex[i], len+1, 0)) return true;
}
}
else
{
if(rt->nex[s[len+1]-'a']!=0 && dfs(rt->nex[s[len+1]-'a'], len+1, 0)) return true;
}
return false;
}
int main()
{
scanf("%d%d", &N, &M);
for(int i=1; i<=N; i++)
{
getchar();
scanf("%s", s+1);
Creat_trie(root, 1);
}
for(int i=1; i<=M; i++)
{
scanf("%s", s+1);
if(dfs(root, 0, 1)) printf("YES\n");
else printf("NO\n");
}
return 0;
}