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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 5048 Accepted Submission(s): 1739

Problem Description

When
you go shopping, you can search in repository for avalible merchandises
by the computers and internet. First you give the search system a name
about something, then the system responds with the results. Now you are
given a lot merchandise names in repository and some queries, and
required to simulate the process.

Input

There
is only one case. First there is an integer P
(1<=P<=10000)representing the number of the merchanidse names in
the repository. The next P lines each contain a string (it's length
isn't beyond 20,and all the letters are lowercase).Then there is an
integer Q(1<=Q<=100000) representing the number of the queries.
The next Q lines each contains a string(the same limitation as foregoing
descriptions) as the searching condition.

Output

For each query, you just output the number of the merchandises, whose names contain the search string as their substrings.

Sample Input

ads

add

ade

adf

adg

adh

adi

adj

adk

adl

aes

Sample Output

我好气呀 g++无限爆内存换c++过了无奈下顺便写了个静态的

题意就是有n个字符串，m个询问问字符串在n个字符串中出现过多少次

比如 abcd 中有a,b,c,d,abcd,bcd,cd,bc,abc...

我们可以将abcd拆成abcd bcd cd d 分别建树，每个字符计算下数字

但abab会导致重复，所以我们设置一下标记，看代码

动态建树，g++爆内存 c++交

 #include<iostream>

 #include<cstdio>

 #include<cmath>

 #include<cstring>

 #include<algorithm>

 #include<queue>

 #include<map>

 #include<set>

 #include<vector>

 #include<cstdlib>

 #include<string>

 #define eps 0.000000001

 typedef long long ll;

 typedef unsigned long long LL;

 using namespace std;

 struct tire{

     int id,num;

     tire *next[];

 };

 tire *root;

 void insert(char *s,int k){

     tire *p,*q;

     p=root;

     int len=strlen(s);

     for(int i=;i<len;i++){

         //cout<<3<<endl;

         int t=s[i]-'a';

         if(p->next[t]==NULL){

             q=(tire *)malloc(sizeof(tire));

             for(int j=;j<;j++)q->next[j]=NULL;

             q->num=;

             q->id=-;

             p->next[t]=q;

         }

         p=p->next[t];

         if(p->id!=k){

             p->id=k;

             p->num++;

         }

     }

 }

 int find(char *s){

     tire *p=root;

     int len=strlen(s);

     for(int i=;i<len;i++){

         int t=s[i]-'a';

         if(p->next[t]==NULL)return ;

         else

             p=p->next[t];

     }

     return p->num;

 }

 int main(){

     int m,n;

     char str[];

     root=(tire *)malloc(sizeof(tire));

     for(int i=;i<;i++)root->next[i]=NULL;

     root->id=-;

     root->num=;

     scanf("%d",&n);

     for(int i=;i<n;i++){

         scanf("%s",str);

         int len=strlen(str);

         for(int j=;j<len;j++){

             //cout<<1<<endl;

             insert(str+j,i);//cout<<2<<endl;

         }

     }

     scanf("%d",&m);

     while(m--){

         scanf("%s",str);

         cout<<find(str)<<endl;

     }

 }

接下来是一个静态的字典树（节省内存） c++ g++都可以过

#include<iostream>

#include<cstdio>

#include<cmath>

#include<cstring>

#include<algorithm>

#include<queue>

#include<map>

#include<set>

#include<vector>

#include<cstdlib>

#include<string>

#define maxnode 500000

#define sigma_size 30

#define eps 0.000000001

typedef long long ll;

typedef unsigned long long LL;

using namespace std;

int ch[maxnode][sigma_size];

int val[maxnode];

int flag[maxnode];

int sz;

void init(){

    memset(ch[],,sizeof(ch[]));

    sz=;

}

int idx(char c){

    return c-'a';

}

void insert(char *s,int k){

    int u=;

    int len=strlen(s);

    for(int i=;i<len;i++){

        int c=idx(s[i]);

        if(ch[u][c]==){

           memset(ch[sz],,sizeof(ch[sz]));

           val[sz]=;

           ch[u][c]=sz++;

        }

        u=ch[u][c];

        if(flag[u]!=k){

            val[u]++;

            flag[u]=k;

        }

    }

}

int find(char *s){

    int u=;

    int len=strlen(s);

    for(int i=;i<len;i++){

        int c=idx(s[i]);

        if(ch[u][c]==)return ;

        u=ch[u][c];

    }

    return val[u];

}

int main(){

    int m,n;

    init();

    memset(flag,-,sizeof(flag));

    //for(int i=0;i<10;i++)cout<<flag[i]<<" ";

    char str[];

    scanf("%d",&n);

    for(int i=;i<n;i++){

        scanf("%s",str);

        int len=strlen(str);

        for(int j=;j<len;j++){

            //cout<<1<<endl;

            insert(str+j,i);//cout<<2<<endl;

        }

    }

    scanf("%d",&m);

    while(m--){

        scanf("%s",str);

        cout<<find(str)<<endl;

    }

}

Source

2009 Multi-University Training Contest 4 - Host by HDU

巴特西

hdu 2846（字典树）

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