[BZOJ 3172] [Tjoi2013] 单词 【AC自动机】

题目链接:BZOJ - 3172

 

题目分析:

  题目要求求出每个单词出现的次数,如果把每个单词都在AC自动机里直接跑一遍,复杂度会很高。

  这里使用AC自动机的“副产品”——Fail树,Fail树的一个性质是,一个字符串出现的次数,就等于以它的结点为根的Fail树中的子树中所有结点的 Cnt 和。

  所以把每个单词插入的时候每个字符都 ++Cnt ,在建 Fail 的时候将结点依次压入一个栈,最后再从栈顶开始弹栈,更新栈顶元素的 Fail 的 Cnt 值,这样就是自叶子节点向上更新了。

  我开始写的时候出现的错误:建 Fail 的时候漏掉了 if (Now -> Child[i] == NULL) Now -> Child[i] = Now -> Fail -> Child[i]; 这句。这样会 RE !

 

代码如下:

  

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <algorithm>
 
using namespace std;
 
const int MaxN = 200 + 5, MaxL = 1000000 + 5, MaxC = 26;
 
int n, l;
 
char Str[MaxL];
 
struct Trie 
{
    int Cnt;
    Trie *Fail, *Child[MaxC];
    void clear() {
        Cnt = 0;
        Fail = NULL;
        for (int i = 0; i < 26; ++i) Child[i] = NULL;
    }
} TA[MaxL], *P = TA, *Root, *Zero, *Pos[MaxN];
 
Trie *NewNode() {
    ++P;
    P -> clear();
    return P;
}
 
void AC_Init() {
    Zero = NewNode();
    Root = NewNode();
    Root -> Fail = Zero;
    for (int i = 0; i < 26; ++i) Zero -> Child[i] = Root;
}
 
Trie *Insert(char *Str, int l) {
    Trie *Now = Root;
    int t;
    for (int i = 0; i < l; ++i) {
        t = Str[i] - 'a';
        if (Now -> Child[t] == NULL) Now -> Child[t] = NewNode();
        Now = Now -> Child[t];
        ++(Now -> Cnt);
    }
    return Now;
}
 
Trie *Q[MaxL], *S[MaxL];
int Head, Tail, Top;
 
void Build_Fail() { 
    Top = 0;
    Head = Tail = 0;
    Q[++Tail] = Root;
    Trie *Now;
    while (Head < Tail) {
        Now = Q[++Head];
        S[++Top] = Now;
        for (int i = 0; i < 26; ++i) {
            if (Now -> Child[i] == NULL) Now -> Child[i] = Now -> Fail -> Child[i];
            else { 
                Now -> Child[i] -> Fail = Now -> Fail -> Child[i];
                Q[++Tail] = Now -> Child[i];
            }
        }
    }
    while (Top) {
        Now = S[Top--];
        if (Now -> Fail != NULL) 
            (Now -> Fail -> Cnt) += (Now -> Cnt);
    }
}
 
int main() 
{
    scanf("%d", &n);
    AC_Init();
    for (int i = 1; i <= n; ++i) {
        scanf("%s", Str);
        l = strlen(Str);
        Pos[i] = Insert(Str, l);
    }
    Build_Fail();
    for (int i = 1; i <= n; ++i) printf("%d\n", Pos[i] -> Cnt);
    return 0;
}

  

  

posted @ 2014-12-10 20:29  JoeFan  阅读(220)  评论(0编辑  收藏  举报