【POJ 4052】Hrinity（AC自动机）

【POJ 4052】Hrinity（AC自动机）

题目是PDF版，地址在：http://poj.org/problem?id=4044

题目大意：多模式串匹配，给出的是经过压缩的串，解压即可。
问文本串有几个模式串能跟他匹配。如果有多个互相包含的模式串，只记录最外层的。

因为说解压前后总长度不超过 5∗105 ，所以全部建出个AC自动机。

然后先用文本串匹配，把匹配到的标记。然后对于每个被标记的模式串，跑一遍，把子串去掉标记。

代码如下：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define Pr pair
#define fread(ch) freopen(ch,"r",stdin)
#define fwrite(ch) freopen(ch,"w",stdout)

using namespace std;
const int INF = 0x3f3f3f3f;
const int msz = 10000;
const int mod = 1e9+7;
const double eps = 1e-8;
const int maxn = 5123456;

struct Node
{
    int next[26],fail,cnt;
} nd[maxn];

bool vis[6666];
int pos[6666];
int id[6666];
char str[maxn];
char ps[maxn];
int len,tp;

int NewNode()
{
    memset(nd[tp].next,0,sizeof(nd[tp].next));
    nd[tp].fail = nd[tp].cnt = 0;
    return tp++;
}

void Add(int root,int c)
{
    int k;
    while(str[len])
    {
        k = str[len]-'A';
        if(!nd[root].next[k]) nd[root].next[k] = NewNode();
        root = nd[root].next[k];
        len++;
    }

    if(!nd[root].cnt) id[c] = nd[root].cnt = c;
    else id[c] = nd[root].cnt;
}

void Build_Fail(int root)
{
    queue <int> q;

    q.push(root);
    int f;

    while(!q.empty())
    {
        root = q.front();
        //printf("%d %d\n",root,nd[root].fail);
        q.pop();
        for(int i = 0; i < 26; ++i)
        {
            if(!nd[root].next[i]) continue;
            if(root) 
            {
                f = nd[root].fail;
                while(f && !nd[f].next[i]) f = nd[f].fail;
                nd[nd[root].next[i]].fail = nd[f].next[i]? nd[f].next[i]: f;
            }
            q.push(nd[root].next[i]);
        }

    }
}

int Search(int root,char *str,int len,int id,int f)
{

    //for(int i = 0; i < len; ++i)
    //  putchar(str[i]);
    //puts("");

    int ans = 0,k;
    for(int i = 0; i < len; ++i)
    {
        k = str[i]-'A';
        while(root && !nd[root].next[k]) root = nd[root].fail;

        if(nd[root].next[k]) root = nd[root].next[k];
        else continue;

        if(f) vis[nd[root].cnt] = f;
        else
        {
            for(int j = root; j; j = nd[j].fail)
            {
                if(nd[j].cnt != id) vis[nd[j].cnt] = f;
            }
        }
    }
    return ans;
}

bool IsChar(char ch)
{
    return 'A' <= ch && ch <= 'Z';
}

void read(char *str)
{
    char ch;
    int pos = 0;

    while((ch = getchar()) && ch == '\n' || ch == EOF || ch == '\r' || ch == ' ');

    while(ch && ch != '\n' && ch != EOF && ch != '\r')
    {
        if(IsChar(ch))
        {
            str[pos++] = ch;
            ch = getchar();
            continue;
        }

        int cnt = 0;
        ch = getchar();
        while(!IsChar(ch))
        {
            cnt = cnt*10+ch-'0';
            ch = getchar();
        }

        while(cnt--) str[pos++] = ch;
        ch = getchar();
        ch = getchar();
    }
    str[pos] = 0;
}

int main()
{
    //fread("");
    //fwrite("");

    int t,n;

    scanf("%d",&t);

    while(t--)
    {
        scanf("%d",&n);

        len = 0;
        tp = 0;
        int root = NewNode();

        for(int i = 1; i <= n; ++i)
        {
            pos[i] = len;
            read(str+len);
            Add(root,i);
        }
        pos[n+1] = len;

        read(ps);

        Build_Fail(root);
        //puts(ps);
        //puts(str);

        memset(vis,0,sizeof(vis));
        Search(root,ps,strlen(ps),0,1);

        for(int i = 1; i <= n; ++i)
        {
            //printf("%d %d\n",id[i],vis[i]);
            if(id[i] == i && vis[i]) Search(root,str+pos[i],pos[i+1]-pos[i],id[i],0);
        }

        int ans = 0;
        for(int i = 1; i <= n; ++i)
            ans += vis[i];

        printf("%d\n",ans);
    }

    return 0;
}