Hat’s Words（字典树）

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

a ahat hat hatword hziee word

 

 

Sample Output

ahat hatword

 

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<cstdio>
 2 #include<iostream>
 3 #include<algorithm>
 4 #include<cstring>
 5 #include<cmath>
 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<len;i++){
21         j=s[i]-'a';
22         if(!ch[k][j]){
23             mem(ch[sz],0);
24         //    val[sz]=0;
25             ch[k][j]=sz++;
26         }
27         k=ch[k][j];
28         word[k]++;
29     }
30     val[k]=1;
31 }
32 bool find(char *s){
33     int len=strlen(s),k=0,j;
34     for(int i=0;i<len;i++){
35         j=s[i]-'a';
36         k=ch[k][j];
37         if(!word[k])return false;
38     }
39     if(val[k])return true;
40     else return false;
41 }
42 int main(){
43     initial();
44     int t=0;
45     char s[110],l[110],r[110];
46     while(~scanf("%s",dt[t]))join(dt[t++]);
47     for(int i=0;i<t;i++){
48         int len=strlen(dt[i]);
49         if(len<=1)continue;
50         for(int j=1;j<len;j++){
51             strcpy(r,dt[i]+j);
52             strcpy(l,dt[i]);
53             l[j]='\0';
54             if(find(l)&&find(r)){
55                 puts(dt[i]);break;
56             }
57         }
58     }
59     return 0;
60 }

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
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<tp;i++){
		for(int j=0;str[i][j];j++){
			strcpy(l,str[i]);
			l[j]='\0';
			strcpy(r,str[i]+j);
	//		printf("**%s %s\n",l,r);
			if(find(l)&&find(r)){
				puts(str[i]);break;
			}
		}
	}
	return 0;
}