因为是模糊匹配,需要用dfs枚举所有情况
坑点 1:有重复的模式
2:字符串匹配完后还可能有*号
样例:
6 1
t
t
t*
t*
t**
t***
输出:0 1 2 3 4 5
代码:
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
using namespace std;
const int maxn = 100005;
int trie[maxn][28];
int val[maxn];
int ans[maxn];
int n, m, tot, anscnt;
vector <int> G[maxn];
int ID(char ch)
{
if (ch == '*') return 26;
if (ch == '?') return 27;
return ch-'a';
}
void insert(char str[],int cnt)
{
int u = 0, id;
int len = strlen(str);
for (int i = 0; i < len; i++)
{
id = ID(str[i]);
if (!trie[u][id])
trie[u][id] = tot++;
u = trie[u][id];
}
val[u] = cnt;
G[u].push_back(cnt);
}
void query(char word[],int u,int pos,int k)
{
int id = 26;
if (trie[u][id])//"*"号必须放在判断结束的前面,因为字符串匹配结束后还可能跟着***
for (int i = pos; i <= k; i++)
query(word,trie[u][id],i,k);
if (pos == k && G[u].size() != 0)
{
int len = G[u].size();
for (int i = 0; i < len; i++)
ans[anscnt++] = G[u][i]-1;
return;
}
id = ID(word[pos]);
if (trie[u][id])
query(word,trie[u][id],pos+1,k);
id = 27;
if (trie[u][id])
query(word,trie[u][id],pos+1,k);
}
void init()
{
memset(val,0,sizeof(val));
memset(trie,0,sizeof(trie));
for (int i = 0; i < maxn; i++)
G[i].clear();
tot = 1;
}
int main()
{
char str[25];
while (scanf("%d%d",&n,&m) != EOF)
{
init();
for (int i = 0; i < n; i++)
{
scanf("%s",str);
insert(str,i+1);
}
while (m--)
{
anscnt = 0;
scanf("%s",str);
int len = strlen(str);
query(str,0,0,len);
if (anscnt == 0)
{
puts("Not match");
continue;
}
sort(ans,ans+anscnt);
for (int i = 0; i < anscnt; i++)
{
if (i != 0 && ans[i] == ans[i-1]) continue;
printf("%d ",ans[i]);
}
puts("");
}
}
return 0;
}