hdu1247 Hat’s Words（字典树）

判断某一个单词是否可以由另外两个单词组合而成，如果是则输出。

把所有单词边读入边插入字典树，树中值同样为某单词是否出现。然后遍历每个单词，再暴力求某单词是否由其他两个组成。由于只有一组数据，所以不存在什么超时爆内存。思路很清晰。

#include <stdio.h>
#include <algorithm>
#include <stdlib.h>
#include <string.h>
#include <iostream>

using namespace std;

typedef long long LL;

const int N = 30;
const int INF = 1e8;

struct Trie
{
int sum;
Trie *next[26];
Trie()
{
sum = 0;
for(int i = 0; i < 26; i ++)
next[i] = 0;
}
}*root;

int ans;

void inserttrie(Trie *p, char *str)
{
int k = 0;
while(str[k] != '\0')
{
int id = str[k] - 'a';
if(p -> next[id] == 0)
{
p -> next[id] = new Trie;
}
p = p -> next[id];
k ++;
}
if(p -> sum == 0) ans ++;
p -> sum ++;//该单词是否出现过
}

int findtrie(Trie *p, char *str)
{
int k = 0, num = 0;
while(str[k] != '\0')
{
int id = str[k] - 'a';
if(p -> next[id])
{
p = p -> next[id];
num = p -> sum;
k ++;
}
else return 0;
}
return num;
}

int main()
{
// freopen("in.txt", "r", stdin);
char s[50005]
, a
, b
;
int countt = 0;
root = new Trie;
while(~scanf("%s", s[countt]))
{
inserttrie(root, s[countt]);
countt ++;
}
for(int i = 0; i < countt; i ++)
{
int len = strlen(s[i]), flag = 0;
for(int j = 0; j < len; j ++)
{
memset(a, 0, sizeof(a));
memset(b, 0, sizeof(b));
int pos1 = 0, pos2 = 0;
for(int l = 0; l < j; l ++)
{
a[pos1 ++] = s[i][l];
}
for(int l = j; l < len; l ++)
{
b[pos2 ++] = s[i][l];
}
if(findtrie(root, a) && findtrie(root, b)) flag = 1;
}
if(flag) printf("%s\n", s[i]);
}
return 0;
}