AC自動(dòng)機(jī)學(xué)習(xí)記錄
參考資料
HDU 2222
題意
給出一些模式串和一個(gè)主串，求主串中能匹配到的模式串?dāng)?shù)量此改。
思路
AC自動(dòng)機(jī)裸題跃巡。
代碼
#include <stdio.h>
#include <string.h>
#include <malloc.h>
#include <queue>
using namespace std;
struct node {
int count;
struct node *next[26];
struct node *fail;
void init() {
for(int i=0; i < 26; ++i)
next[i] = NULL;
count = 0;
fail = NULL;
}
} *root;
void insert(char str[]) {
node *p = root;
int len = strlen(str);
for(int k=0; k < len; ++k) {
int pos = str[k]-'a';
if(p->next[pos] == NULL) {
p->next[pos] = new node;
p->next[pos]->init();
p = p->next[pos];
}
else
p = p->next[pos];
}
p->count++;
}
void getfail() {
node *p = root, *son, *temp;
queue<node*> que;
que.push(p);
while(!que.empty()){
temp = que.front(); que.pop();
for(int i=0; i < 26; ++i){
son = temp->next[i];
if(son != NULL){
if(temp == root)
son->fail = root;
else{
p = temp->fail;
while(p){
if(p->next[i]){
son->fail=p->next[i];
break;
}
p = p->fail;
}
if(!p)
son->fail = root;
}
que.push(son);
}
}
}
}
void query(char str[]) {
int len = strlen(str), cnt=0;
node *p,*temp;
p = root;
for (int i=0; i < len; ++i) {
int pos = str[i]-'a';
while(!p->next[pos] && p!=root)
p = p->fail;
p = p->next[pos];
if(!p) p = root;
temp = p;
while (temp != root) {
if(temp->count >= 0) {
cnt += temp->count;
temp->count = -1;
}
else break;
temp = temp->fail;
}
}
printf("%d\n", cnt);
}
char str[1000000+100];
int main() {
int cas,n;
scanf("%d",&cas);
while(cas--) {
root=new node;
root->init();
root->fail = NULL;
scanf("%d",&n);
int i;
for(i=0;i<n;i++) {
scanf("%s", str);
insert(str);
}
getfail();
scanf("%s", str);
query(str);
}
return 0;
}
HDU 2896
題意
給出一些模式串，再給出一些主串，求每個(gè)主串中能匹配到的模式串的編號(hào)次慢，以及能匹配到模式串的主串?dāng)?shù)量岩馍。
思路
本來(lái)是想用上面 HDU 2222 的模板改的，發(fā)現(xiàn)會(huì)MLE伙菊，就重新找了一個(gè)模板败玉。
這個(gè)模板是開(kāi)一個(gè)node數(shù)組，通過(guò)next和fail都通過(guò)下標(biāo)來(lái)指示位置占业，
數(shù)據(jù)量一定時(shí)節(jié)省空間但是不能根據(jù)數(shù)據(jù)量自動(dòng)拓展绒怨。
代碼
#include <cstdio>
#include <cstring>
#include <cmath>
#include <iostream>
#include <algorithm>
#include <queue>
using namespace std;
const int kind = 128;
const int s_maxn = 200+7;
const int t_maxn = 10000+7;
const int maxn = 500+7;
struct node {
int fail, flag, next[kind];
void init() {
for(int i=0; i < kind; ++i)
next[i] = 0;
fail = -1, flag = 0;
}
} tb[100007];
int web_cnt, n, m, cur;
char s[s_maxn], t[t_maxn];
void insert(const char str[], int id) {
int temp = 0;
for(int k=0; str[k]; ++k) {
int pos = str[k];
if(tb[temp].next[pos] == 0) {
tb[++cur].init();
tb[temp].next[pos] = cur;
}
temp = tb[temp].next[pos];
}
tb[temp].flag = id;
}
void getfail() {
int p, son;
queue<int> que;
que.push(0);
while(!que.empty()){
int temp = que.front(); que.pop();
for(int i=0; i < kind; ++i){
son = tb[temp].next[i];
if (!son) continue;
p = tb[temp].fail;
while (~p && !tb[p].next[i])
p = tb[p].fail;
if (!~p)
tb[son].fail = 0;
else
tb[son].fail = tb[p].next[i];
que.push(son);
}
}
}
void query(char str[], int id) {
int ans[maxn] = {}, cnt = 0;
int p = 0, temp;
for (int i=0; str[i]; ++i) {
int pos = str[i];
if (tb[p].next[pos])
p = tb[p].next[pos];
else {
temp = tb[p].fail;
while (~temp && !tb[temp].next[pos])
temp = tb[temp].fail;
if (p == -1) p = 0;
else p = tb[temp].next[pos];
}
if(tb[p].flag) {
ans[tb[p].flag] = 1;
++cnt;
temp = tb[p].fail;
while (temp && !ans[tb[temp].flag]) {
ans[tb[temp].flag] = 1;
temp = tb[temp].fail;
++cnt;
}
}
}
if (cnt == 0) return;
++web_cnt;
printf("web %d:", id);
for (int i = 1; i <= n; ++i)
if (ans[i]) printf(" %d", i);
puts("");
}
void solve() {
web_cnt = cur = 0;
tb[0].init();
for (int i = 1; i <= n; ++i) {
scanf("%s", s);
insert(s, i);
}
getfail();
int m; scanf("%d", &m);
for (int i = 1; i <= m; ++i) {
scanf("%s", t);
query(t, i);
}
printf("total: %d\n", web_cnt);
}
int main(){
#ifdef LOCAL
freopen("in.txt", "r", stdin);
#endif
while (~scanf("%d", &n))
solve();
return 0;
}
