hdu

统计难题

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131070/65535 K (Java/Others)
Total Submission(s): 35486    Accepted Submission(s): 13269

Problem Description

Ignatius最近遇到一个难题,老师交给他很多单词(只有小写字母组成,不会有重复的单词出现),现在老师要他统计出以某个字符串为前缀的单词数量(单词本身也是自己的前缀).

 

Input

输入数据的第一部分是一张单词表,每行一个单词,单词的长度不超过10,它们代表的是老师交给Ignatius统计的单词,一个空行代表单词表的结束.第二部分是一连串的提问,每行一个提问,每个提问都是一个字符串.

注意:本题只有一组测试数据,处理到文件结束.

 

Output

对于每个提问,给出以该字符串为前缀的单词的数量.

 

Sample Input

banana band bee absolute acm ba b band abc

 

Sample Output

2 3 1 0

 

Author

Ignatius.L

 

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#include <map>
#include <set>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <queue>
#include <iostream>
#include <stack>
#include <cmath>
#include <string>
#include <vector>
#include <cstdlib>
//#include <bits/stdc++.h>
//#define LOACL
#define space " "
using namespace std;
//typedef long long LL;
typedef __int64 Int;
typedef pair<int, int> paii;
const int INF = 0x3f3f3f3f;
const double ESP = 1e-5;
const double PI = acos(-1.0);
const int MAXN = 1000000 + 10;
char str[1000], ask[1000];
int ch[MAXN][40], word[MAXN], val[MAXN];
int sz;
void init() {
    sz = 1;
    memset(ch[0], 0, sizeof(ch[0]));
    memset(word, 0, sizeof(word));
}
int idx(char x) {return x - 'a';}
void insert_str(char *s) {
    int i, j, l = strlen(s);
    int u = 0;
    for(i = 0; i < l; i++) {
        int c = idx(s[i]);
        if(!ch[u][c]) {
            memset(ch[sz], 0, sizeof(ch[sz]));
            val[sz] = 0; ch[u][c] = sz++;
        }
        u = ch[u][c];
        word[u]++;
    }
    val[sz] = 1;
}
int find_str(char s[]) {
    int i, j, l = strlen(s);
    int u = 0;
    for(i = 0; i < l; i++) {
        int c = idx(s[i]);
        if(!ch[u][c]) return 0;
        u = ch[u][c];
    }
    return word[u];
}
int main() {
    init();
    while(gets(str), str[0]) insert_str(str);
    while(scanf("%s", ask) != EOF) {
        printf("%d
", find_str(ask));
    }
    return 0;
}