Light OJ Substring Frequency (II)

[align=center]1427 - Substring Frequency (II)[/align]

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Time Limit: 5 second(s)

Memory Limit: 128 MB

A string is a finite sequence of symbols that are chosen from an alphabet. In this problem you are given a string T and n queries each with a string Pi,
where the strings contain lower case English alphabets only. You have to find the number of times Pi occurs as a substring of T.

Input

Input starts with an integer T (≤ 10), denoting the number of test cases.

Each case starts with a line containing an integer n (1 ≤ n ≤ 500). The next line contains the string T (1 ≤ |T| ≤ 106). Each of the next n lines
contains a string Pi (1 ≤ |Pi| ≤ 500).

Output

For each case, print the case number in a single line first. Then for each string Pi, report the number of times it occurs as a substring of T in a single line.

Sample Input	Output for Sample Input
2 5 ababacbabc aba ba ac a abc 3 lightoj oj light lit	Case 1: 2 3 1 4 1 Case 2: 1 1 0

分析：AC自动机模版

#include<cstdio>
#include<queue>
#include<cstring>
using namespace std;
char str[1000002];
char s[502][502];
int tot,cnt;
struct trie{
int flag,fail,next[26],index;
void init(){
flag=0;fail=index=-1;
for(int i=0;i<26;i++)next[i]=0;
}
}q[500002];
void insert(char *S){
int t=0,i=0,c;
while(S[i]){
c=S[i]-'a';
if(!q[t].next[c]){
q[++tot].init();
q[t].next[c]=tot;
}
t=q[t].next[c];
i++;
}
q[t].flag=1;
q[t].index=cnt;
}
void get_fail(){
int i;
queue<int> Q;
Q.push(0);
while(!Q.empty()){
int p=Q.front();Q.pop();
for(i=0;i<26;i++){
if(q[p].next[i]){
int a=q[p].fail,b=q[p].next[i];
while(a!=-1&&!q[a].next[i]) a=q[a].fail;
if(a==-1) q[b].fail=0;
else q[b].fail=q[a].next[i];
Q.push(b);
}
}
}
}
int ans[502];
void match(){
int c,i=0,p,t=0;
while(str[i]){
c=str[i]-'a';
p=t;
while(p!=-1&&!q[p].next[c]) p=q[p].fail;
if(p==-1) t=0;
else t=q[p].next[c];
p=t;
while(p!=0){
ans[q[p].index]+=q[p].flag;
//q[p].flag=0;
p=q[p].fail;
}
i++;
}
int j;
for(i=0;i<cnt;i++){
if(ans[i]>0)printf("%d\n",ans[i]);
else {
for(j=0;j<cnt;j++)
if(i!=j&&strcmp(s[i],s[j])==0&&ans[j]>0) {printf("%d\n",ans[j]);break;}
if(j==cnt)puts("0");
}
}
}
int main(){
int t,n,ca;
scanf("%d",&t);
for(ca=1;ca<=t;ca++){
scanf("%d",&n);
scanf("%s",str);
tot=0;q[0].init();
cnt=0;
while(n--){
scanf("%s",s[cnt]);
insert(s[cnt]);
cnt++;
}
printf("Case %d:\n",ca);
get_fail();
memset(ans,0,sizeof(ans));
match();
}
return 0;
}