字符串-Trie字典树
Skyzhou666 · · 算法·理论
Trie
增查模板
#include <iostream>
#define MAXN 1001
using namespace std;
int son[MAXN][26], cnt[MAXN], idx;
void insert(string s)
{
int p = 0; //指向当前节点
for(int x = 0; x < s.length(); x++)
{
int u = s[x] - 'a';
if(!son[p][u]) son[p][u] = ++idx;
p = son[p][u];
}
cnt[p]++;
}
int query(string s)
{
int p = 0;
for(int x = 0; x < s.length(); x++)
{
int u = s[x] - 'a';
if(!son[p][u]) return 0;
p = son[p][u];
}
return cnt[p];
}
int main()
{
int n, m;
string s;
cin >> n >> m;
for(int x = 0; x < n; x++)
{
cin >> s;
insert(s);
}
for(int x = 0; x < m; x++)
{
cin >> s;
cout << query(s) << endl;
}
return 0;
}
这篇kingwzun的blog讲的不错
P2922 USACO Secret Message G
基本上就是模板,但考虑到题目询问的是有关前缀的问题,要稍微小改一下
#include <iostream>
#include <cstring>
#define MAXN 500001
using namespace std;
int inp[MAXN], son[MAXN][2], cnt[MAXN], idx, wrd[MAXN];
void trie_insert(int i)
{
int p = 0;
for(int x = 0; x < i; x++)
{
int u = inp[x];
if(!son[p][u]) son[p][u] = ++idx;
p = son[p][u];
cnt[p]++;
}
wrd[p]++;
}
int trie_query(int i)
{
int p = 0, ans = 0;
for(int x = 0; x < i; x++)
{
int u = inp[x];
if(!son[p][u]) return ans;
p = son[p][u];
ans += wrd[p];
}
return ans-wrd[p]+cnt[p];
}
int main()
{
int n, m, i;
cin >> m >> n;
while(m--)
{
cin >> i;
for(int x = 0; x < i; x++) cin >> inp[x];
trie_insert(i);
}
while(n--)
{
cin >> i;
for(int x = 0; x < i; x++) cin >> inp[x];
cout << trie_query(i) << endl;
}
return 0;
}