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;
}