一.P8306 【模板】字典树
题目思路:字典树的板子题,熟练写出insert函数(建树),以及query函数(查询)即可.
代码实现:
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define N 3000005
#define P 131
ll f1[N];
string s;
char ss[N];
ll n, m, num, ans, t, id;
int cnt[N],ch[N][126];
typedef pair<char, ll>pii;
void clear()
{
for (int i = 0; i <= id; i++) {
cnt[i] = 0;
for (int j = 0; j <= 126; j++) {
ch[i][j] = 0;
}
}
id = 0;
}
void insert(string s) {
int p=0;
for (int i = 0; i <= s.size(); i++) {
int j = (int)s[i];
if (!ch[p][j]) ch[p][j] = ++id;
p = ch[p][j];
cnt[p]++;
}
}
ll find(string s) {
int p = 0;
for (int i = 0; i < s.size(); i++) {
int j = (int)s[i];
if (!ch[p][j]) return 0;
p = ch[p][j];
}
return cnt[p];
}
int main()
{
cin >> t;
while (t--) {
clear();
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> s;
insert(s);
}
for (int i = 1; i <= m; i++) {
cin >> s;
cout << find(s) << endl;
}
}
return 0;
}二.P1481 魔族密码
题目思路:因为是记录最长词链,我们只要在query函数中,使ans+=cnt[p],即将每个节点的单词个数进行累加即可,insert函数不变.
代码实现:
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define N 3000005
#define P 131
char s[N];
int ch[N][27],cnt[N];
int n, m, k, num, sum;
void insert(char s[])
{
int p = 0;
for (int i = 0; s[i]; i++) {
int q = s[i] - 'a' + 1;
if (!ch[p][q]) ch[p][q] = ++num;
p = ch[p][q];
}
cnt[p]++;
}
int query(char s[]) {
int p = 0, ans = 0;
for (int i = 0; s[i]; i++) {
int q = s[i] - 'a' + 1;
if (!ch[p][q]) return 0;
p = ch[p][q];
ans += cnt[p];
}
return ans;
}
int main()
{
cin >> n;
for (int i = 1; i <= n; i++) {
cin >> s;
insert(s);
sum = max(query(s), sum);
}
cout << sum << endl;
return 0;
}三.P2580 于是他错误的点名开始了
题目思路:先把所有的人名字存起来，通过insert函数，这样可以建立从名字到数的映射，标记为1，然后扫一遍，如果为1，输出OK，同时将cnt数组++，如果遇到>1，输出REPEAT，如果为0，说明没有，输出WRONG
代码实现:
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define N 3000005
#define P 131
char s[N];
int ch[N][27],cnt[N];
int n, m, k, num, sum;
void insert(char s[])
{
int p = 0;
for (int i = 0; s[i]; i++) {
int q = s[i] - 'a' + 1;
if (!ch[p][q]) ch[p][q] = ++num;
p = ch[p][q];
}
cnt[p]=1;
}
int query(char s[]) {
int p = 0, ans = 0;
for (int i = 0; s[i]; i++) {
int q = s[i] - 'a' + 1;
if (!ch[p][q]) return 0;
p = ch[p][q];
}
if (!cnt[p]) return 0;
if (cnt[p] == 1) {
cnt[p]++;
return 1;
}
return 2;
}
int main()
{
cin >> n;
memset(cnt, 0, sizeof(cnt));
for (int i = 1; i <= n; i++) {
cin >> s;
insert(s);
}
cin >> m;
for (int i = 1; i <= m; i++) {
cin >> s;
int k = query(s);
if (k == 0)
cout << "WRONG" << endl;
if (k == 1)
cout << "OK" << endl;
if (k == 2)
cout << "REPEAT" << endl;
}
return 0;
}四.P2922 [USACO08DEC] Secret Message G
题目思路:和魔族密码这道题有点像,但是我们要维护两个数组,即一个ans数组记录经过该节点的次数,另一个cnt数组记录以该节点结束的次数,在query时,不断累加ans数组的值,最后再sum - cnt[p] + ans[p](因为在最后时应该把以当前结束的次数去掉).
代码实现:
#include<bits/stdc++.h>
using namespace std;
typedef unsigned long long ull;
typedef long long ll;
#define N 3000005
#define P 131
ll s[N];
ll ch[N][3],cnt[N],ans[N];
int n, m, k, num=1, sum;
void insert(ll s[])
{
ll p = 1;
for (int i = 1; i <= k; i++) {
ll q = s[i];
if (!ch[p][q]) ch[p][q] = ++num;
p = ch[p][q];
ans[p]++;
}
cnt[p]++;
}
int query(ll s[]) {
ll p = 1;
sum = 0;
for (int i = 1; i <= k; i++) {
ll q = s[i];
if (!ch[p][q]) return sum;
p = ch[p][q];
sum += cnt[p];
}
return sum - cnt[p] + ans[p];
}
int main()
{
cin >> n >> m;
for (int i = 1; i <= n; i++) {
cin >> k;
for (int j = 1; j <= k; j++) cin >> s[j];
insert(s);
}
for (int i = 1; i <= m; i++) {
cin >> k;
for (int j = 1; j <= k; j++) cin >> s[j];
cout << query(s) << endl;
}
return 0;
}
