HDU 1247 Hat’s Words // Trie, 枚举

题目描述

解题思路

参考代码

//**********************************************
//  Author: @xmzyt1996
//  Date:   2015-10-15
//  Name:   HDU 1247.cpp
//**********************************************
#include <cstdio>
#include <cmath>
#include <cctype>
#include <cstring>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <string>
#include <bitset>
#include <vector>
#include <stack>
#include <queue>
#include <map>
#define clr(a, b) memset(a, b, sizeof(a))

using namespace std;

const int MAX_N = 50010;
const int MAX_V = 110;

struct Trie {
Trie* next[26];
int flag;
}node[MAX_N<<5];

int num;

void init() {
num = 0;
memset(node, 0, sizeof(node));
}

int getid(char c) { return c - 'a'; }

void insert(char* s) {
Trie* head = &node[0];
while (*s) {
int id = getid(*s++);
if ((head->next[id]) == NULL)
head->next[id] = &node[++num];
head = head->next[id];
}
head->flag = 1;
}

bool search(char* s) {
Trie* head = &node[0];
while (*s) {
int id = getid(*s++);
if ((head->next[id]) == NULL)
return false;
head = head->next[id];
}
return head->flag;
}

char word[MAX_N][MAX_V], s1[MAX_V], s2[MAX_V];

int main() {
int k = 0;
while (~scanf("%s", word[k])) insert(word[k++]);
for (int i = 0; i < k; ++i) {
int len = strlen(word[i]);
for (int j = 1; j < len; ++j) { // 枚举每个单词所有的组合
clr(s1, 0); clr(s2, 0);
strncpy(s1, &word[i][0], j);  // 前半段
strncpy(s2, &word[i][j], len-j);  // 后半段
if(search(s1) && search(s2)) { // 若分解的两个单词都存在，则满足
puts(word[i]);
break;
}
}
}
return 0;
}