String
Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)Total Submission(s): 1053 Accepted Submission(s): 339
Problem Description
Bob has a dictionary with N words in it.
Now there is a list of words in which the middle part of the word has continuous letters disappeared. The middle part does not include the first and last character.
We only know the prefix and suffix of each word, and the number of characters missing is uncertain, it could be 0. But the prefix and suffix of each word can not overlap.
For each word in the list, Bob wants to determine which word is in the dictionary by prefix and suffix.
There are probably many answers. You just have to figure out how many words may be the answer.
Now there is a list of words in which the middle part of the word has continuous letters disappeared. The middle part does not include the first and last character.
We only know the prefix and suffix of each word, and the number of characters missing is uncertain, it could be 0. But the prefix and suffix of each word can not overlap.
For each word in the list, Bob wants to determine which word is in the dictionary by prefix and suffix.
There are probably many answers. You just have to figure out how many words may be the answer.
Input
The first line of the input gives the number of test cases T; T test cases follow.
Each test case contains two integer N and Q, The number of words in the dictionary, and the number of words in the list.
Next N line, each line has a string Wi, represents the ith word in the dictionary ( 0<|Wi|≤100000)
Next Q line, each line has two string Pi , Si, represents the prefix and suffix of the ith word in the list ( 0<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤100000)
All of the above characters are lowercase letters.
The dictionary does not contain the same words.
Limits
T≤5
0<N,Q≤100000
∑Si+Pi≤500000
∑Wi≤500000
Each test case contains two integer N and Q, The number of words in the dictionary, and the number of words in the list.
Next N line, each line has a string Wi, represents the ith word in the dictionary ( 0<|Wi|≤100000)
Next Q line, each line has two string Pi , Si, represents the prefix and suffix of the ith word in the list ( 0<|Pi|,|Si|≤100000,0<|Pi|+|Si|≤100000)
All of the above characters are lowercase letters.
The dictionary does not contain the same words.
Limits
T≤5
0<N,Q≤100000
∑Si+Pi≤500000
∑Wi≤500000
Output
For each test case, output Q lines, an integer per line, represents the answer to each word in the list.
Sample Input
1 4 4 aba cde acdefa cdef a a cd ef ac a ce f
Sample Output
2 1 1 0
Source
Recommend
题目大意：给你n个字符串，然后有q个询问，每个询问都给一个字符串的前驱和后缀，要求输出所给出的这个前驱和后缀可能在多少个字符串中出现。
解题思路：因为题目上给出的是前缀和后缀，所以我们在建立字典树时的顺序为str0,strn-1,str2,strn-2··· ···strn-1,str0;设前缀为s 后缀为t，则在查询时查询的字符串也为s0, tn-1,s1, tn-2,前后缀长度不等用*补齐。（如ac f，构造后的字符串为 afc*）.
字典树需要维护3个值，这个节点的后继节点child，经过这个节点的字符串有多少个cnt，经过这个节点的字符串的长度len(用动态数组来存)。
在查询时因为要防止 字符串为 aaa 前缀为 aa 后缀为 aa 这种情况出现，所以我们需要对每个节点中vector的数值排序，去掉长度小于前缀+后缀的值，最后cnt的值就可能是当前这个前缀和后缀可能组成字符串的个数了。
这里还有一个要注意的地方就是当出现 * 的时候 * 后边所有的节点都可能是我们要匹配的答案，所以所有节点都要记录下来留作处理。这里我用连个队列来处理我们的节点。一个队列用来存放父亲节点，一个队列用来存放孩子节点，两个队列交换使用，实现对所有可能节点的计数。
注意：这个题的内存给的比较大，而时间比较少，如果用链表的方式来建树可能会超时，这里我给出链表建树和数组建树两种方法。链表式的方法只是用来助理解的，正确的还是要用数组来存。
链表建树：
#include <cstdio>
#include <iostream>
#include <string.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>
#include <vector>
#define N 100000+10
using namespace std;
struct Trie
{
vector<int> len;
int cnt;
Trie *child[26];
Trie()///构造函数，完成每个节点的初始化
{
cnt = 0;
len.clear();
for(int i=0; i<26; i++)
child[i]=NULL;
}
};
Trie *root,*current,*temp;
void insert(char s[],int lens)
///建树时要维护cnt（表示经过这个节点的字符串有多少个）和动态数组len（保存经过这个节点的字符串的长度len）
{
current = root;
for(int i=0; i<strlen(s); i++)
{
if(current->child[s[i]-'a']==NULL)
{
temp = new Trie;
current->child[s[i]-'a'] = temp;
current = current->child[s[i]-'a'];
current->len.push_back(lens);
current->cnt = 1;
}
else
{
current = current->child[s[i]-'a'];
current->cnt++;
current->len.push_back(lens);
}
}
}
void dfs(Trie *root)
{
for(int i=0; i<26; i++)
if(root->child[i]!=NULL)
dfs(root->child[i]);
sort(root->len.begin(),root->len.end());
}
int search(char s[],int lenn,int ret)
{
queue<Trie*> Q[2];
Q[0].push(root);
int ans = 0,tmp = 0;
for(int i=0;i < lenn;i++){
tmp = 1-tmp;
while(!Q[1-tmp].empty()){
///每次改变队列，实现当前节点在一个队列，当前节点的子节点在一个队列，不漏掉任何一个可能的节点
current = Q[1-tmp].front();
Q[1-tmp].pop();
if(s[i] == '*'){
for(int j = 0; j < 26; j++)
{
if(current->child[j]!=NULL)
Q[tmp].push(current->child[j]);
}
}
else{
if(current->child[s[i]-'a']!=NULL)
Q[tmp].push(current->child[s[i]-'a']);
}
}
}
while(!Q[tmp].empty()){
///因为可能出现*后边有很多个节点的情况，所以可能有多个满足条件的节点，每个可能满足条件的节点都要计算
Trie *t= new Trie;
t = Q[tmp].front();
Q[tmp].pop();
int tt = lower_bound(t->len.begin(),t->len.end(),ret) - t->len.begin();
///lower_bound返回的是大于或等于ret的第一个元素的位置，再减去元素首位置，就是这个数组中有多少个元素比ret小
ans -= tt;
ans += t->cnt;
}
return ans;
}
void del(Trie *root)
{
for(int i=0;i<26;i++)
if(root->child[i]!=NULL)
del(root->child[i]);
delete(root);
}
int main()
{
int T;
cin>>T;
while(T--)
{
root = new Trie;
int n,q;
char str[N],st[2*N],s[N],t[N];
scanf("%d %d",&n,&q);
for(int i=0; i<n; i++)
{
scanf("%s",str);
int L = strlen(str);
for(int i=0; i<L; i++)///st用来保存处理后的字符串即str0，strn-1，str1，strn-2······
{
st[i*2] = str[i];
st[i*2+1] = str[L-1-i];
}
st[2*L] = '\0';
insert(st,L);///构造字典树
}
dfs(root);
for(int i=0; i<q; i++)
{
scanf("%s %s",s,t);
int len1 = strlen(s);
int len2 = strlen(t);
int ret = len1+len2;
int len = max(len1,len2);
for(int j= 0; j < len; j ++)///将前后缀串按前边的约定来构造
{
if(j < len1) st[j*2] = s[j];
else st[j*2] = '*';
if(j < len2) st[j*2+1] = t[len2-1-j];
else st[j*2+1] = '*';
}
len*=2;
st[len] = '\0';
printf("%d\n",search(st,len,ret));
}
del(root);//防止内存超
}
return 0;
}
数组建树：
#include<bits/stdc++.h>
using namespace std;
const int N = 5e5+100;
char s[N],t[N],st[N*2];
int n,q;
struct Tire{
int L,root,net[N*2][26],ed[2*N];
vector<int> G[2*N];
int newnode(){
for(int i= 0;i < 26;i ++) net[L][i] = -1;
ed[L] = 0;
return L++;
}
void init(){
L = 0;
for(int i= 0;i < 2*N;i ++)
G[i].clear();
root = newnode();
}
void build(char *s,int len){
int lens = strlen(s);
int now = root;
for(int i=0 ;i < lens;i ++){
int id = s[i]-'a';
if(net[now][id] == -1){
net[now][id] = newnode();
}
now = net[now][id];
G[now].push_back(len);
ed[now]++;
}
}
void dfs(int x){
sort(G[x].begin(),G[x].end());
for(int i = 0;i < 26;i ++){
if(net[x][i] != -1) dfs(net[x][i]);
}
}
}tire;
int main(){
int T;
cin >> T;
while(T--){
tire.init();
scanf("%d %d",&n,&q);
for(int i= 1;i <= n;i ++){
scanf("%s",&s);
int len = strlen(s);
for(int i= 0;i < len;i ++){
st[i*2] = s[i];
st[i*2+1] = s[len-1-i];
}
st[len*2] = '\0';
tire.build(st,len);
}
tire.dfs(0);
for(int i= 1;i <= q;i ++){
scanf("%s %s",s,t);
int len1 = strlen(s);
int len2 = strlen(t);
int ret = len1+len2;
int len = max(len1,len2);
for(int j= 0;j < len;j ++){
if(j < len1) st[j*2] = s[j];
else st[j*2] = '*';
if(j < len2) st[j*2+1] = t[len2-1-j];
else st[j*2+1] = '*';
}
len*=2;
st[len] = '\0';
queue<int> q[2];
int tmp = 0;
q[0].push(0);
int ans = 0;
for(int j= 0;j < len;j ++){
tmp = 1-tmp;
int id = st[j]-'a';
while(!q[1-tmp].empty()){
int now = q[1-tmp].front();
q[1-tmp].pop();
if(st[j] == '*'){
for(int k = 0;k < 26;k ++){
if(tire.net[now][k] != -1){
q[tmp].push(tire.net[now][k]);
}
}
}
else{
if(tire.net[now][id] != -1){
q[tmp].push(tire.net[now][id]);
}
}
}
}
while(!q[tmp].empty()){
int now = q[tmp].front();
q[tmp].pop();
int cnt = lower_bound(tire.G[now].begin(),tire.G[now].end(),ret) - tire.G[now].begin();
ans -= cnt;
ans += tire.ed[now];
}
printf("%d\n",ans);
}
}
return 0;
}