题目链接:

​http://acm.hdu.edu.cn/showproblem.php?pid=2896​



题目大意:

给你N个模式串(编号为1~N)。接下来给你M个主串。问:主串中出现的模式串的编号。


思路:

和HDU2222一样。都是求文本串(主串)中出现的模式串。不同的是这道题要求输出的是模式

串的编号。用Val[]来保存模式串的编号id。同时注意字符为ASCII码可见字符,即32~127共

95个。其实定义95就可以了,我这里定义了128,不过没什么影响。


AC代码:


#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<queue>
using namespace std;
const int MAXN = 210*500;
const int SIZE = 128;

struct Trie
{ //Next[i][j]表示字典中i节点的字符为j的儿子节点编号
int Next[MAXN][SIZE];
int Fail[MAXN]; //失配数组
int Val[MAXN]; //出现模式串的id
int root,L;
int NewNode()
{
for(int i = 0; i < SIZE; ++i)
Next[L][i] = -1;
Val[L++] = -1;
return L-1;
}
void Init()
{
L = 0;
root = NewNode();
}
void Insert(char *s,int id)
{
int len = strlen(s);
int now = root;
for(int i = 0; i < len; ++i)
{
if(Next[now][s[i]] == -1)
Next[now][s[i]] = NewNode();
now = Next[now][s[i]];
}
Val[now] = id;
}
void Build()
{
queue<int> Q;
Fail[root] = root;
for(int i = 0; i < SIZE; ++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 < SIZE; ++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]);
}
}
}
bool vis[510];
bool Query(char *buf,int n,int id)
{
int len = strlen(buf);
int now = root;
memset(vis,false,sizeof(vis));
bool flag = false;
for(int i = 0; i < len; ++i)
{
now = Next[now][buf[i]];
int temp = now;
while(temp != root)
{
if(Val[temp] != -1)
{
vis[Val[temp]] = true;
flag = true;
}
temp = Fail[temp];
}
}
if(!flag)
return false;
printf("web %d:",id);
for(int i = 1; i <= n; ++i)
if(vis[i])
printf(" %d",i);
printf("\n");
return true;
}
};

char Buf[10010];
Trie AC;

int main()
{
int N,M;
while(~scanf("%d",&N))
{
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;
}