解题思路：
对AC自动机算法的理解和应用，要注意的是，题目中的字符是128个ASCII码，而不是常用的26个英文字符，因此子节点的数目为128.
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
#include <queue>
#include <string.h>
#include <stack>
#include <algorithm>
using namespace std;
struct Trie
{
int next[210*500][128],fail[210*500],end[210*500];
int root,L;
int newnode()
{
for(int i = 0;i < 128;i++)
next[L][i] = -1;
end[L++] = -1;
return L-1;
}
void init()
{
L = 0;
root = newnode();
}
void insert(char buf[], int id)
{
int len = strlen(buf);
int now = root;
for(int i = 0;i < len;i++)
{
if(next[now][buf[i]] == -1)
next[now][buf[i]] = newnode();
now = next[now][buf[i]];
}
end[now] = id;
}
void build()
{
queue<int>Q;
fail[root] = root;
for(int i = 0;i < 128;i++)
if(next[root][i] == -1)
next[root][i] = root;
else
{
fail[next[root][i]] = root;
Q.push(next[root][i]);
}
while( !Q.empty() )
{
int now = Q.front();
Q.pop();
for(int i = 0;i < 128;i++)
if(next[now][i] == -1)
next[now][i] = next[fail[now]][i];
else
{
fail[next[now][i]]=next[fail[now]][i];
Q.push(next[now][i]);
}
}
}
int vis[510];
int query(char buf[],int n, int id)
{
int len = strlen(buf);
int now = root;
int flag = 0;
memset(vis, 0, sizeof(vis));
for(int i = 0;i < len;i++)
{
now = next[now][buf[i]];
int temp = now;
while( temp != root )
{
if(end[temp] != -1)
{
vis[end[temp]] = 1;
flag = 1;
}
temp = fail[temp];
}
}
if(!flag) return 0;
printf("web %d:", id);
for(int i=1;i<=n;i++) if(vis[i])
printf(" %d", i);
printf("\n");
return 1;
}
};
char buf[10010];
Trie ac;
int main()
{
int n, m;
while(scanf("%d", &n)!=EOF)
{
ac.init();
for(int i=1;i<=n;i++)
{
scanf("%s", buf);
ac.insert(buf, i);
}
ac.build();
int ans = 0;
scanf("%d", &m);
for(int i=1;i<=m;i++)
{
scanf("%s", buf);
if(ac.query(buf, n, i))
ans++;
}
printf("total: %d\n", ans);
}
return 0;
}
原文：http://blog.csdn.net/moguxiaozhe/article/details/46314747