lightoj 1427 Substring Frequency (||) (AC自动机)

1427 - Substring Frequency (II)

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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 P_i, where the strings contain lower case English alphabets only. You have to find the number of times P_i 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 P_i(1 ≤ |P_i| ≤ 500).

Output

For each case, print the case number in a single line first. Then for each string P_i, 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

Notes

1.      Dataset is huge, use faster I/O methods.

2.       If S is a string then |S| denotes the length of S.

题目链接：http://lightoj.com/volume_showproblem.php?problem=1427

题目大意：问单词在给定的段落中出现的次数

题目分析：AC自动机，注意单词可能重复，插单词的时候标记一下，把idx转到其第一次出现的坐标，本题动态new的话，不delete会mle，delete会超时

下面给出三种常见姿势和一个神奇姿势:

1.普通静态指针型 (3956 ms)

#include <cstdio>#include <cstring>int const MAXN = 1e6 + 5;int const MAXM = 505;char s[MAXM], t[MAXN];int tail, head, sz;int ans[MAXM], cg[MAXM];struct NODE{int cnt;int id;int idx;NODE *fail;NODE *next[26];}*q[MAXM * MAXM], *root, *f, node[MAXM * MAXM];inline void Newnode(NODE * &x){x = &node[sz];x -> id = sz ++;x -> cnt = 0;x -> idx = 0;x -> fail = NULL;memset(x -> next, NULL, sizeof(x -> next));}inline void Init(){sz = 0;head = tail = 0;f = &node[0];f -> id = 0;f -> cnt = 0;f -> idx = 0;f -> fail = NULL;memset(f -> next, NULL, sizeof(f -> next));Newnode(root);}inline void Insert(NODE *p, char *s, int id){int len = strlen(s);for(int i = 0; i < len; i++){int idx = s[i] - 'a';if(p -> next[idx] == NULL)Newnode(p -> next[idx]);p = p -> next[idx];}if(p -> cnt)cg[id] = p -> id; else{p -> id = id;p -> cnt = 1;}}inline void Build_AC(NODE *root){q[tail ++] = root;while(tail != head){NODE *p = q[head ++];for(int i = 0; i < 26; i++){if(p -> next[i] != NULL){if(p == root)p -> next[i] -> fail = root;else{NODE *tmp = p -> fail;while(tmp != NULL){if(tmp -> next[i] != NULL){p -> next[i] -> fail = tmp -> next[i];   break;}tmp = tmp -> fail;}if(tmp == NULL)p -> next[i] -> fail = root;}q[tail ++] = p -> next[i];}}}}inline void Query(NODE *root, char *t){NODE *p = root;int len = strlen(t);for(int i = 0; i < len; i++){int idx = t[i] - 'a';while(p != root && p -> next[idx] == NULL)p = p -> fail;if(p -> next[idx] == NULL)p = root;elsep = p -> next[idx];NODE *tmp = p;while(tmp != root){ans[tmp -> id] += tmp -> cnt;tmp = tmp -> fail;}}}int main(){int T;scanf("%d", &T);for(int ca = 1; ca <= T; ca++){memset(ans, 0, sizeof(ans));memset(cg, -1, sizeof(cg));Init();int n;scanf("%d", &n);scanf("%s", t);for(int i = 0; i < n; i++){scanf("%s", s);Insert(root, s, i);}Build_AC(root);Query(root, t);printf("Case %d:\n", ca);for(int i = 0; i < n; i++){if(cg[i] != -1)printf("%d\n", ans[cg[i]]);elseprintf("%d\n", ans[i]);}}}

2.Tire图优化静态指针型 (4000ms) 。。反而更慢

#include <cstdio>#include <cstring>int const MAXN = 1e6 + 5;int const MAXM = 505;char s[MAXM], t[MAXN];int tail, head, sz;int ans[MAXM], cg[MAXM];struct NODE{int cnt;int id;int idx;NODE *fail;NODE *next[26];}*q[MAXM * MAXM], *root, *f, node[MAXM * MAXM];inline void Newnode(NODE * &x){x = &node[sz];x -> id = sz ++;x -> cnt = 0;x -> idx = 0;x -> fail = NULL;memset(x -> next, NULL, sizeof(x -> next));}inline void Init(){sz = 0;head = tail = 0;f = &node[0];f -> id = 0;f -> cnt = 0;f -> idx = 0;f -> fail = NULL;memset(f -> next, NULL, sizeof(f -> next));Newnode(root);}inline void Insert(NODE *p, char *s, int id){int len = strlen(s);for(int i = 0; i < len; i++){int idx = s[i] - 'a';if(p -> next[idx] == NULL)Newnode(p -> next[idx]);p = p -> next[idx];}if(p -> cnt)cg[id] = p -> id; else{p -> id = id;p -> cnt = 1;}}inline void Build_AC(NODE *root){q[tail ++] = root;while(tail != head){NODE *p = q[head ++];for(int i = 0; i < 26; i++){if(p -> next[i] != NULL){if(p == root)p -> next[i] -> fail = root;elsep -> next[i] -> fail = p -> fail -> next[i];q[tail ++] = p -> next[i];}else{if(p == root)p -> next[i] = root;elsep -> next[i] = p -> fail -> next[i];}}}}inline void Query(NODE *root, char *t){NODE *p = root;int len = strlen(t);for(int i = 0; i < len; i++){int idx = t[i] - 'a';while(p != root && p -> next[idx] == NULL)p = p -> fail;if(p -> next[idx] == NULL)p = root;elsep = p -> next[idx];NODE *tmp = p;while(tmp != root){ans[tmp -> id] += tmp -> cnt;tmp = tmp -> fail;}}}int main(){int T;scanf("%d", &T);for(int ca = 1; ca <= T; ca++){memset(ans, 0, sizeof(ans));memset(cg, -1, sizeof(cg));Init();int n;scanf("%d", &n);scanf("%s", t);for(int i = 0; i < n; i++){scanf("%s", s);Insert(root, s, i);}Build_AC(root);Query(root, t);printf("Case %d:\n", ca);for(int i = 0; i < n; i++){if(cg[i] != -1)printf("%d\n", ans[cg[i]]);elseprintf("%d\n", ans[i]);}}}

3.静态数组Trie图优化 (2668 ms)

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;int const MAXN = 1e6 + 5;int const MAXM = 505 * 505 ;char t[MAXN], s[MAXM];int n, ans[MAXM], cg[MAXM];struct AC{int next[MAXM][26], fail[MAXM], end[MAXM], id[MAXM];int root, tot;inline int Newnode(){for(int i = 0; i < 26; i++)next[tot][i] = -1;end[tot] = 0;fail[tot] = 0;return tot ++;}inline void Init(){tot = 0;root = Newnode();}inline void Insert(char *s, int idx){int len = strlen(s);int now = root;for(int i = 0; i < len; i++){int idx = s[i] - 'a';if(next[now][idx] == -1)next[now][idx] = Newnode();now = next[now][idx];}if(end[now])cg[idx] = id[now];else{end[now] = 1;id[now] = idx;}}inline void Build(){queue <int> q;q.push(root);while(!q.empty()){int now = q.front();q.pop();for(int i = 0; i < 26; i++){if(next[now][i] == -1){if(now == root)next[now][i] = root;elsenext[now][i] = next[fail[now]][i];}else{if(now == root)fail[next[now][i]] = root;elsefail[next[now][i]] = next[fail[now]][i];q.push(next[now][i]);}}}}inline void Query(char *s){int now = root;int len = strlen(s);for(int i = 0; i < len; i++){int idx = s[i] - 'a';while(now != root && next[now][idx] == -1)now = fail[now];now = next[now][idx];int tmp = now;while(tmp != root){ans[id[tmp]] += end[tmp];tmp = fail[tmp];}}}}ac;int main(){int T;scanf("%d", &T);for(int ca = 1; ca <= T; ca++){memset(cg, -1, sizeof(cg));memset(ans, 0, sizeof(ans));ac.Init();scanf("%d", &n);scanf("%s", t);for(int i = 0; i < n; i++){scanf("%s", s);ac.Insert(s, i);}ac.Build();ac.Query(t);printf("Case %d:\n", ca);for(int i = 0; i < n; i++){if(cg[i] != -1)printf("%d\n", ans[cg[i]]);elseprintf("%d\n", ans[i]);}}}

4.静态数组Trie图优化，并用一个数组记录所有fail的点，累计的时候只看母串上的和有fail的点，注意fail的值是从深的地方往浅的地方回溯的 (700ms)

#include <cstdio>#include <cstring>#include <queue>#include <algorithm>using namespace std;int const MAXN = 1e6 + 5;int const MAXM = 505 * 505;char t[MAXN], s[MAXM];int ans[MAXM], pos[MAXM];int n;struct AC{int next[MAXM][26], fail[MAXM];int st[MAXM];int root, tot, top;inline int Newnode(){memset(next[tot], -1, sizeof(next[tot]));ans[tot] = 0;fail[tot] = 0;return tot ++;}inline void Init(){top = 0;tot = 0;root = Newnode();}inline void Insert(char *s, int id){int len = strlen(s);int now = root;for(int i = 0; i < len; i++){int idx = s[i] - 'a';if(next[now][idx] == -1)next[now][idx] = Newnode();now = next[now][idx];}pos[id] = now;}inline void Build(){queue <int> q;q.push(root);while(!q.empty()){int now = q.front();q.pop();for(int i = 0; i < 26; i++){if(next[now][i] == -1){if(now == root)next[now][i] = root;elsenext[now][i] = next[fail[now]][i];}else{if(now == root)fail[next[now][i]] = root;elsefail[next[now][i]] = next[fail[now]][i];q.push(next[now][i]);st[++ top] = next[now][i];}}}}inline void Query(char *s){int now = root;int len = strlen(s);for(int i = 0; i < len; i++){int idx = s[i] - 'a';now = next[now][idx];ans[now] ++;}while(top){int p = st[top];ans[fail[p]] += ans[p];top --;}}}ac;int main(){int T;scanf("%d", &T);for(int ca = 1; ca <= T; ca++){ac.Init();scanf("%d", &n);scanf("%s", t);for(int i = 1; i <= n; i++){scanf("%s", s);ac.Insert(s, i);}ac.Build();ac.Query(t);printf("Case %d:\n", ca);for(int i = 1; i <= n; i++)printf("%d\n", ans[pos[i]]);}}