HDU-1251-统计难题(字典树)

链接:

https://vjudge.net/problem/HDU-1251

题意:

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).

思路:

直接考虑字典数, 每个节点记录经过的次数,次数就是前缀数目.

代码:

#include <iostream>
#include <cstdio>
#include <cstring>
#include <vector>
//#include <memory.h>
#include <queue>
#include <set>
#include <map>
#include <algorithm>
#include <math.h>
#include <stack>
#include <string>
#include <assert.h>
#include <iomanip>
#define MINF 0x3f3f3f3f
using namespace std;
typedef long long LL;
const int INF = 1e9;
const int MAXN = 1e6+10;

int cnt, n;
char word[100];

struct Node
{
    int Ends;
    int Pass;
    int Next[30];
    void Init()
    {
        Ends = 0;
        Pass = 0;
        memset(Next, 0, sizeof(Next));
    }
}Trie[MAXN];

void Insert(char* val)
{
    int root = 0;
    int len = strlen(val);
    for (int i = 0;i < len;i++)
    {
        if (Trie[root].Next[val[i]-'a'] == 0)
        {
            Trie[root].Next[val[i]-'a'] = ++cnt;
            Trie[cnt].Init();
        }
        root = Trie[root].Next[val[i]-'a'];
        Trie[root].Pass++;
    }
    Trie[root].Ends++;
}

int Query(char *val)
{
    int root = 0;
    int len = strlen(val);
    for (int i = 0;i < len;i++)
    {
        if (Trie[root].Next[val[i]-'a'] == 0)
            return 0;
        root = Trie[root].Next[val[i]-'a'];
    }
    return Trie[root].Pass;
}

int main()
{
    while (gets(word))
    {
        if (word[0] == '\0')
            break;
        Insert(word);
    }
    while (~scanf("%s", word))
    {
        printf("%d\n", Query(word));
    }

    return 0;
}

来源：https://www.cnblogs.com/YDDDD/p/11578092.html