标签:style blog http io ar color os 使用 sp
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; }
[BZOJ 3172] [Tjoi2013] 单词 【AC自动机】
标签:style blog http io ar color os 使用 sp
原文地址:http://www.cnblogs.com/F-Magician/p/4156236.html