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：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 }