BZOJ3172 单词 Fail树

题目大意：求一篇论文中每个单词分别在论文中出现多少次。

本题用AC自动机太慢，应该用Fail树将AC自动机中所有的Fail指针反向得到一个新树，这就是Fail树。对长度为x的字符串a和长度为y的字符串b，如果a是b的子串，则a可能与位于b[0,a],b[0,a+1],b[0,a+2]...b[0,y]中的后缀相等。根据fail指针的定义，只要沿着反向Fail边走，走到的节点所代表的字符串必然存在与a（前缀）相等的后缀。因此，一遍DFS，返回加上子节点的总Cnt值的当前节点的Cnt值，即可。注意，Trie树中，有些节点是多个字符串公用的，因此每次构造Trie树时，都要对每个节点的Cnt++，以等价于此处存在多个字符串。

#include <cstdio>
#include <cstring>
#include <cassert>
#include <algorithm>
#include <cmath>
#include <queue>
#include <vector>
using namespace std;

const int MAX_CHAR = 26, MAX_LEN = 1e6 + 10, MAX_STR = 210;

struct FailTree
{
#define Root _nodes[0]
#define Org(x) x - 'a'

    struct Node;
    struct Edge;

    struct Node
    {
        Node *Next[MAX_CHAR], *Fail;
        int Cnt;
        Edge *Head;
        Node() :Cnt(0), Fail(NULL), Head(NULL) { memset(Next, NULL, sizeof(Next)); }
    };
    vector<Node*> _nodes, Tail;

    struct Edge
    {
        Node *To;
        Edge *Next;
        Edge(Node *to, Edge *next):To(to),Next(next){}
    };
    vector<Edge*> _edges;

    FailTree()
    {
        _nodes.push_back(new Node());
    }

    void AddEdge(Node *from, Node *to)
    {
        Edge *e = new Edge(to, from->Head);
        from->Head = e;
        _edges.push_back(e);
    }

    Node *BuildTrie(char *s)
    {
        int len = strlen(s);
        Node *cur = Root;
        for (int i = 0; i < len; i++)
        {
            if (!cur->Next[Org(s[i])])
                _nodes.push_back(cur->Next[Org(s[i])] = new Node());
            cur = cur->Next[Org(s[i])];
            cur->Cnt++;
        }
        return cur;
    }

    void Insert(char *s)
    {
        Tail.push_back(BuildTrie(s));
    }

    void SetFail()
    {
        static queue<Node*> q;
        q.push(Root);
        while (!q.empty())
        {
            Node *cur = q.front();
            q.pop();
            for (int i = 0; i < MAX_CHAR; i++)
            {
                if (cur->Next[i])
                {
                    Node *temp = cur->Fail;
                    while (temp)
                    {
                        if (temp->Next[i])
                        {
                            cur->Next[i]->Fail = temp->Next[i];
                            AddEdge(temp->Next[i], cur->Next[i]);
                            break;
                        }
                        temp = temp->Fail;
                    }
                    if (!temp)
                    {
                        cur->Next[i]->Fail = Root;
                        AddEdge(Root, cur->Next[i]);
                    }
                    q.push(cur->Next[i]);
                }
            }
        }
    }

    int Dfs(Node *u)
    {
        for (Edge *e = u->Head; e; e = e->Next)
            u->Cnt += Dfs(e->To);
        return u->Cnt;
    }
}g;

int main()
{
#ifdef _DEBUG
    freopen("c:\\noi\\source\\input.txt", "r", stdin);
#endif
    int tot;
    char s[MAX_LEN];
    scanf("%d", &tot);
    for(int i=0; i<tot; i++)
    {
        scanf("%s", s);
        g.Insert(s);
    }
    g.SetFail();
    g.Dfs(g.Root);
    for (int i = 0; i < tot; i++)
        printf("%d\n", g.Tail[i]->Cnt);
    return 0;
}

 或者不用反向Fail指针也可以，站在后缀上去找其所包含的前缀。这样编程复杂度低一些。

#include <cstdio>
#include <cstring>
#include <vector>
#include <queue>
#include <cassert>
#include <cmath>
#include <algorithm>
using namespace std;

const int MAX_CHAR = 26, MAX_NODE = 5e5 + 1, MAX_LEN = 1e6 + 1;

struct Node
{
    int Sum, Id, Cnt;
    Node *Fail;
    Node *Next[MAX_CHAR];
}Nodes[MAX_NODE];
int Nodes_Cnt = 1;
char P[MAX_LEN];
Node *WordNode[MAX_NODE];

int Ord(char c)
{
    return c - 'a';
}

Node *NewNode()
{
    return ++Nodes_Cnt + Nodes;
}

Node *Root()
{
    return Nodes + 1;
}

void BuildTrie(char *s, int id)
{
    Node *cur = Root();
    int len = strlen(s);
    for (int i = 0; i < len; i++)
    {
        if (cur->Next[Ord(s[i])])
            cur = cur->Next[Ord(s[i])];
        else
            cur = cur->Next[Ord(s[i])] = NewNode();
    }
    cur->Sum++;
    cur->Id = id;
    WordNode[id] = cur;
}

void SetFail()
{
    queue<Node*> q;
    q.push(Root());
    while (!q.empty())
    {
        Node *cur = q.front();
        q.pop();
        for (int i = 0; i < MAX_CHAR; i++)
        {
            if (cur->Next[i])
            {
                Node *temp = cur->Fail;
                while (temp)
                {
                    if (temp->Next[i])
                    {
                        cur->Next[i]->Fail = temp->Next[i];
                        break;
                    }
                    temp = temp->Fail;
                }
                if (!temp)
                {
                    cur->Next[i]->Fail = Root();
                }
                q.push(cur->Next[i]);
            }
        }
    }
}

int Dfs1(Node *cur)
{
    int cnt = cur->Sum;
    for (int i = 0; i < MAX_CHAR; i++)
        if (cur->Next[i])
            cnt += Dfs1(cur->Next[i]);
    for (Node *temp = cur; temp != Root(); temp = temp->Fail)
        if (temp->Sum)
            temp->Cnt+=cnt;
    //cur->Cnt += cnt;
    return cnt;
}

int main()
{
    //freopen("c:\\noi\\source\\input.txt", "r", stdin);
    int totP;
    scanf("%d", &totP);
    for (int i = 0; i < totP; i++)
    {
        scanf("%s", P);
        BuildTrie(P, i);
    }
    SetFail();
    Dfs1(Root());
    for (int i = 0; i < totP; i++)
        printf("%d\n", WordNode[i]->Cnt);
    return 0;
}