POJ-4052-ac自动机

来源：互联网发布：linux 重启防火墙编辑：程序博客网时间：2024/05/01 20:12
先把匹配上的串都标记。然后把每个匹配上的串的子串变成非结束。
直接会超时。优化找失败指针的地方。
#include <iostream>#include <cstdio>#include <cmath>#include <string>#include <cstring>#include <queue>#include <algorithm>using namespace std;const int maxn = 250009;char str1[5100009], str2[5100009]; char s[2509][1109];char ss[1109];int vis[maxn];int cnt, n;struct trie{    trie *ch[26];    trie *fail;    int end, key;    void init()    {        memset(ch, NULL, sizeof(ch));        fail = NULL;        key = 0;    }}*root, t[maxn];void insert(char *s, int id) //正常的插入{    trie *p = root;    for(int i=0; s[i]; i++)    {        int k = s[i]-'A';        if(p->ch[k]==NULL)        {            t[cnt].init();            p->ch[k] = &t[cnt++];        }        p = p->ch[k];    }    p->end = id;    p->key = 1;}void reinsert(char *a) //将其子串删掉{    trie *p = root;    for(int i=0; a[i]; i++)    {        int k = a[i]-'A';        while(p->ch[k]==NULL && p!=root) p=p->fail;        p = p->ch[k];        p = (p==NULL)? root : p;        trie *tmp = p;        while(tmp!=root && tmp->key!=-3)        {            if(abs(tmp->key)==1 && vis[tmp->end])            {                vis[tmp->end] = 0;            }            tmp->key = -3;            tmp = tmp->fail;        }    }}queue<trie*>q;void build_fail(){    q.push(root);    root->fail = root;    while(!q.empty())    {        trie *p = q.front();        q.pop();        for(int k=0; k<26; k++)        if(p->ch[k]!=NULL)        {            q.push(p->ch[k]);            if(p==root)            {                p->ch[k]->fail = root;                continue;            }            trie *tmp = p->fail;            while( tmp!=root && tmp->ch[k]==NULL)                tmp = tmp->fail;            if(tmp->ch[k]) p->ch[k]->fail = tmp->ch[k];            else p->ch[k]->fail = root;        }    }}void query(char *a) //标记出现过的所有串{    trie *p = root;    for(int i=0; a[i]; i++)    {        int k = a[i]-'A';        while(p->ch[k]==NULL && p!=root) p=p->fail;        p = p->ch[k];        p = (p==NULL)? root : p;        trie *tmp = p;        while(tmp!=root && tmp->key>=0)        {            if(tmp->key==1)            {                vis[tmp->end] = 1;                tmp->key = -1;            }            else tmp->key = -2;            tmp = tmp->fail;        }    }}int requery(char *a) {    int ans = 0;    trie *p = root;    for(int i=0; a[i]; i++)    {        int k = a[i]-'A';        while(p->ch[k]==NULL && p!=root) p=p->fail;        p = p->ch[k];        p = (p==NULL)? root : p;        trie *tmp = p;        while(tmp!=root && tmp->key!=-4)        {            if(abs(tmp->key)==1) ans++;            tmp->key = -4;            tmp = tmp->fail;        }    }    return ans;}int main(){    int tt, i, j, k;    scanf("%d", &tt);    while(tt--)    {        cnt = 1;        t[0].init();        root = &t[0];        scanf("%d", &n);        for(i = 1; i<=n; i++)        {            scanf("%s", ss);            for(k=0, j=0; ss[k]; k++)            {                if(ss[k]>='A' && ss[k]<='Z')                {                    s[i][j] = ss[k];                    j++;                }                else                {                    k++;                    int kk = ss[k]-'0';                    k++;                    while(ss[k]>='0'&&ss[k]<='9') kk = kk*10+ ss[k++]-'0';                    while(kk--) s[i][j++] = ss[k];                    k++;                }            }            s[i][j] = '\0';            insert(s[i], i);        }        build_fail();        scanf("%s", str1);        for(i=0, j=0; str1[i]; i++)        {            if(str1[i]>='A' && str1[i]<='Z')            {                str2[j] = str1[i];                j++;            }            else            {                i++;                int k = str1[i]-'0';                i++;                while(str1[i]>='0'&&str1[i]<='9') k = k*10+ str1[i++]-'0';                while(k--) str2[j++] = str1[i];                i++;            }        }        str2[j] = '\0';        memset(vis, 0, sizeof(vis));        query(str2);        for(int i=1; i<=n; i++)        if(vis[i])        {            reinsert( s[i] );            insert( s[i], i );        }        int ans = requery( str2 );        printf("%d\n", ans);    }return 0;}