博客
关于我
Hat’s Words(字典树)
阅读量:620 次
发布时间:2019-03-13

本文共 2202 字,大约阅读时间需要 7 分钟。

Hat’s Words

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 11314    Accepted Submission(s): 4041

Problem Description
A hat’s word is a word in the dictionary that is the concatenation of exactly two other words in the dictionary.
You are to find all the hat’s words in a dictionary.
 

 

Input
Standard input consists of a number of lowercase words, one per line, in alphabetical order. There will be no more than 50,000 words.
Only one case.
 

 

Output
Your output should contain all the hat’s words, one per line, in alphabetical order.
 

 

Sample Input
aahathathatwordhzieeword
 

 

Sample Output
ahathatword
 
题解:判断单词是否是由两个单词合并得到,加了个val判断是否为一个完整的单词;本来还以为是只与hat结合呐,神经的只减去hat,谁知道,自己想错了。。。是两个单词的结合;
代码:
1 #include
2 #include
3 #include
4 #include
5 #include
6 using namespace std; 7 #define mem(x,y) memset(x,y,sizeof(x)) 8 const int INF=0x3f3f3f3f; 9 const double PI=acos(-1.0);10 const int MAXN=50010;11 int ch[MAXN][30],word[MAXN],val[MAXN];12 char dt[50010][110];13 int sz;14 void initial(){15 sz=1;16 mem(ch[0],0);mem(word,0);mem(val,0);17 }18 void join(char *s){19 int len=strlen(s),k=0,j;20 for(int i=0;i

 

#include
#include
#include
#include
#include
using namespace std;typedef long long LL;#define mem(x,y) memset(x,y,sizeof(x))#define SI(x) scanf("%d",&x)#define PI(x) printf("%d",x)#define P_ printf(" ")const int INF=0x3f3f3f3f;const int MAXN=1000010;//要开的足够大 int ch[MAXN][30];int word[MAXN];char str[50010][110];int val[MAXN]; int sz;int N;char l[110],r[110];void insert(char *s){ int k=0,j; for(int i=0;s[i];i++){ j=s[i]-'a'; if(!ch[k][j]){ mem(ch[sz],0); ch[k][j]=sz++; } k=ch[k][j]; word[k]++; } val[k]=1;}bool find(char *s){ int k=0,j; for(int i=0;s[i];i++){ j=s[i]-'a'; k=ch[k][j]; if(!word[k])return false; } if(val[k]!=1)return false; return true;}int main(){ int tp=0; sz=1; mem(ch[0],0);mem(val,0);mem(word,0); while(~scanf("%s",str[tp]))insert(str[tp++]);// printf("%d\n",tp); for(int i=0;i

  

转载地址:http://aueaz.baihongyu.com/

你可能感兴趣的文章
node防xss攻击插件
查看>>
noi 1996 登山
查看>>
noi 7827 质数的和与积
查看>>
NOI-1.3-11-计算浮点数相除的余数
查看>>
noi.ac #36 模拟
查看>>
NOI2010 海拔(平面图最大流)
查看>>
NOIp2005 过河
查看>>
NOIP2011T1 数字反转
查看>>
NOIP2014 提高组 Day2——寻找道路
查看>>
noip借教室 题解
查看>>
NOIP模拟测试19
查看>>
NOIp模拟赛二十九
查看>>
Vue3+element plus+sortablejs实现table列表拖拽
查看>>
Nokia5233手机和我装的几个symbian V5手机软件
查看>>
non linear processor
查看>>
Non-final field ‘code‘ in enum StateEnum‘
查看>>