Danganronpa

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 318    Accepted Submission(s): 176

Problem Description

给你n个A串,m个B串,对每个A串,询问,这些B串们在该A串中一共出现过多少次

Input

样例个数

n m

接下来n个A串

接下来m个B串

Output

如问题描述,对每个A输出...

Sample Input
1
5 6
orz
sto
kirigiri
danganronpa
ooooo
o
kyouko
dangan
ronpa
ooooo
ooooo
 
Sample Output
1
1
0
3
7
 



Source
 
解题:AC自动机自动AC
 
 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
struct trie {
int word[],fail,cnt;
void init() {
fail = cnt = ;
memset(word,-,sizeof word);
}
} dic[maxn];
int tot;
char str[];
void insertWord(int root,char *s) {
for(int i = ; s[i]; i++) {
int k = s[i]-'a';
if(dic[root].word[k] == -) {
dic[++tot].init();
dic[root].word[k] = tot;
}
root = dic[root].word[k];
}
++dic[root].cnt;
}
queue<int>q;
void build(int root) {
while(!q.empty()) q.pop();
q.push(root);
while(!q.empty()) {
int u = q.front();
q.pop();
for(int i = ; i < ; i++) {
if(dic[u].word[i] == -) continue;
if(u == ) dic[dic[u].word[i]].fail = ;
else {
int v = dic[u].fail;
while(v && dic[v].word[i] == -) v = dic[v].fail;
if(dic[v].word[i] == -) dic[dic[u].word[i]].fail = ;
else dic[dic[u].word[i]].fail = dic[v].word[i];
}
q.push(dic[u].word[i]);
}
}
}
int query(int root,char *s) {
int i,ans = ;
for(i = ; s[i]; i++) {
int k = s[i]-'a';
while(root && dic[root].word[k] == -)
root = dic[root].fail;
if(dic[root].word[k] != -) {
int v = dic[root].word[k];
while(v) {
ans += dic[v].cnt;
v = dic[v].fail;
}
root = dic[root].word[k];
}
}
return ans;
}
char FK[][];
int main() {
int n,m,t;
scanf("%d",&t);
while(t--) {
dic[tot = ].init();
scanf("%d%d",&n,&m);
for(int i = ; i < n; ++i)
scanf("%s",FK[i]);
while(m--) {
scanf("%s",str);
insertWord(,str);
}
build();
for(int i = ; i < n; ++i)
printf("%d\n",query(,FK[i]));
}
return ;
}

这才是AC自动机的正确使用方法,Trie图

 #include <bits/stdc++.h>
using namespace std;
const int maxn = ;
struct Trie{
int ch[maxn][],fail[maxn],cnt[maxn],tot;
int newnode(){
memset(ch[tot],,sizeof ch[tot]);
fail[tot] = cnt[tot] = ;
return tot++;
}
void init(){
tot = ;
newnode();
}
void insert(char *str,int root = ){
for(int i = ; str[i]; ++i){
if(!ch[root][str[i]-'a'])
ch[root][str[i]-'a'] = newnode();
root = ch[root][str[i]-'a'];
}
++cnt[root];
}
void build(int root = ){
queue<int>q;
for(int i = ; i < ; ++i)
if(ch[root][i]) q.push(ch[root][i]);
while(!q.empty()){
root = q.front();
q.pop();
for(int i = ; i < ; ++i)
if(ch[root][i]){
fail[ch[root][i]] = ch[fail[root]][i];
cnt[ch[root][i]] += cnt[ch[fail[root]][i]];
q.push(ch[root][i]);
}else ch[root][i] = ch[fail[root]][i];
}
}
int query(char *str,int ret = ,int root = ){
for(int i = ; str[i]; ++i){
int x = root = ch[root][str[i]-'a'];
ret += cnt[x];
}
return ret;
}
}ac;
char FK[][],str[];
int main(){
int kase,n,m;
scanf("%d",&kase);
while(kase--){
ac.init();
scanf("%d%d",&n,&m);
for(int i = ; i < n; ++i)
scanf("%s",FK[i]);
while(m--){
scanf("%s",str);
ac.insert(str);
}
ac.build();
for(int i = ; i < n; ++i)
printf("%d\n",ac.query(FK[i]));
}
return ;
}

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