链接：
http://acm.hdu.edu.cn/showproblem.php?pid=2846
题目：
Repository
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1657 Accepted Submission(s): 605
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.
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.
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
20 ad ae af ag ah ai aj ak al ads add ade adf adg adh adi adj adk adl aes 5 b a d ad s
Sample Output
0 20 11 11 2
题目大意：
给P个单词， 然后有Q个询问， 每个问给一个单词t， 然后问在P个单词中，t是其字串的共有多少个。
分析与总结：
Trie树的应用。
对于每个单词， 把他的所有字串插入字符串，并在相应的位置+1. 注意一个单词如果有多个相同的字串，那这个字串也只能加一次。所有节点要多一个变量用来标记是否已经加过。
这个过程只需要枚举单词的开始位置，然后插入Trie树即可。
然后查询直接找出相应单词在Trie树中记录的数量即可。
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
const int KIND = 26;
const int MAXN = 5000000;
int cnt_node;
struct node{
int flag; // 标记是否已经加过了
int cnt; // 记录以这个单词为字串的有多少个商品
node* next[KIND];
void init(){
flag=-10;
cnt=0;
memset(next, 0, sizeof(next));
}
}Heap[MAXN];
inline node* new_node(){
Heap[cnt_node].init();
return &Heap[cnt_node++];
}
inline void insert(node *root, char *str, int flag){
for(char *p=str; *p; ++p){
int ch=*p-'a';
if(root->next[ch]==NULL)
root->next[ch]=new_node();
root = root->next[ch];
if(root->flag!=flag){
root->flag=flag;
++root->cnt;
}
}
}
inline int get_count(node* root, char* str){
for(char *p=str; *p; ++p){
int ch=*p-'a';
if(root->next[ch]==NULL)
return 0;
root = root->next[ch];
}
return root->cnt;
}
int main(){
int p;
char str[25];
scanf("%d",&p);
// Trie init
cnt_node=0;
node *root = new_node();
while(p--){
scanf("%s",str);
char tmp[25];
int len=strlen(str);
for(int i=0; i<len; ++i){
insert(root, str+i, p);
}
}
scanf("%d",&p);
while(p--){
scanf("%s",str);
printf("%d\n", get_count(root, str));
}
return 0;
}
—— 生命的意义，在于赋予它意义士。
原创 http://blog.csdn.net/shuangde800 ， By
D_Double (转载请标明)