zstu-2437 English word

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2437: English word

Time Limit: 1 Sec Memory Limit: 64 MB
Submit: 2580 Solved: 1174

Description

You still are worried about reading acm English problem, let me tell you a kind of very good method of Memorising Words, the root memory method,.the most interesting of the method is the prefix root. Now there are some English words to remembeand he knows some prefix root;He want to know how many words that can be remembered using each the prefix .in other words,how many words that contain the prefix ,Now to acmer you for help, can you help him?

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Input

The input consists of T test cases. The number of them (T) is given on the first line of the input file. Each test case begins with a line containing a single integer number N that indicates the number of words (1 <= N <= 10000). Then exactly N lines follow, each containing a single word. Each word contains at least two and at most 10 lowercase characters, that means only letters ‘a‘ through ‘z‘ will appear in the word. The same word may appear several times in the list. Then follow a line containing a single integer number M that indicates the number of Words prefix question (0 <= M<= 10000). Then exactly M lines follow, each containing a single Words prefix.

Output

For the given input data, output exact M line, each line contain a ingle integer indicates the answer of each query.

Sample Input

1
4
preview
predict
premier
press
3
pre
press
pree

Sample Output

4
1
0

HINT

Source

style_luo

用来练一下字典树模板，当年用快排+二分做的也很开心

#include<bits/stdc++.h>
using namespace std;
#define ll long long
typedef struct node {
    int num;
    node *next[30];
    node() {
        memset(next, 0, sizeof(next));
        num = 0;
    }
}Trie;
char op[32], s[32];
void Insert(node *root, char *s) {
    node *p = root;
    for(int i = 0; s[i]; i++) {
        int x = s[i] - ‘a‘;
        if(p -> next[x] == NULL) p -> next[x] = new node;
        p = p -> next[x];
        p -> num++;
    }
}
int Search(node *root, char *s) {
    node *p = root;
    for(int i = 0; s[i]; i++) {
        int x = s[i] - ‘a‘;
        if(p -> next[x] == NULL) return 0;
        p = p -> next[x];
    }
    return p -> num;
}
 
void Delete(node *root, char *s, int cnt) {
    node *p = root;
    for(int i = 0; s[i]; i++) {
        int x = s[i] - ‘a‘;
        p = p -> next[x];
        p -> num -= cnt;
    }
    for(int i = 0; i < 30; i++) p -> next[i] = 0;
}
 
int main() {
    int t;
    scanf("%d", &t);
    while(t--) {
        Trie *root = new node;
        int n;
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%s", s);
            Insert(root, s);
        }
        scanf("%d", &n);
        for(int i = 1; i <= n; i++) {
            scanf("%s", s);
            printf("%d\n", Search(root, s));
        }
    }
}

#include<bits/stdc++.h>
using namespace std;
string ch[11111];
string pre[11111];
int cnt=0;
 
void jiancha_right(int val2,string pr2)
{
    if(val2 >= 0 && ch[val2].compare(0, pr2.size(), pr2) == 0)
    {
        cnt++;
        jiancha_right(val2+1,pr2);
    }
    else return;
}
 
void jiancha_left(int val2,string pr2)
{
    if(val2 >= 0 && ch[val2].compare(0, pr2.size(), pr2) == 0)
    {
        cnt++;
        jiancha_left(val2-1,pr2);
    }
    else return;
}
 
int bbsearch(string pr,int m1,int n1)
{
    int val;
    while(m1<n1)
    {
        val=m1+(n1-m1)/2;
        //printf("%s %s %d %d %d\n",ch[val],pr,m1,n1,val);
        if(ch[val].compare(0, pr.size(), pr)==0)
        {
            cnt++;
            jiancha_left(val-1,pr);
            jiancha_right(val+1,pr);
            break;
        }
        else if(pr < ch[val]) n1=val;
        else m1=val+1;
    }
    return cnt;
}
 
int main()
{
    int t,m,n;
    char s[1];
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&m);
        for(int i=0;i<m;i++)
        {
            cin >> ch[i];
        }
        scanf("%d",&n);
        for(int i=0;i<n;i++)
        {
            cin >> pre[i];
        }
 
        sort(ch, ch + m);
        int ans;
 
        for(int i=0;i<n;i++)
        {
            cnt=0;
            ans=bbsearch(pre[i],0,m);
            printf("%d\n",ans);
        }
 
    }
}

zstu-2437 English word

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原文地址：http://www.cnblogs.com/lonewanderer/p/5697989.html