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


[align=left]Problem Description[/align]
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.

[align=left]Input[/align]
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.

[align=left]Output[/align]
Your output should contain all the hat’s words, one per line, in alphabetical order.

[align=left]Sample Input[/align]

a
ahat
hat
hatword
hziee
word

[align=left]Sample Output[/align]

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：

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<cmath>
using namespace std;
#define mem(x,y) memset(x,y,sizeof(x))
const int INF=0x3f3f3f3f;
const double PI=acos(-1.0);
const int MAXN=50010;
int ch[MAXN][30],word[MAXN],val[MAXN];
char dt[50010][110];
int sz;
void initial(){
sz=1;
mem(ch[0],0);mem(word,0);mem(val,0);
}
void join(char *s){
int len=strlen(s),k=0,j;
for(int i=0;i<len;i++){
j=s[i]-'a';
if(!ch[k][j]){
mem(ch[sz],0);
//    val[sz]=0;
ch[k][j]=sz++;
}
k=ch[k][j];
word[k]++;
}
val[k]=1;
}
bool find(char *s){
int len=strlen(s),k=0,j;
for(int i=0;i<len;i++){
j=s[i]-'a';
k=ch[k][j];
if(!word[k])return false;
}
if(val[k])return true;
else return false;
}
int main(){
initial();
int t=0;
char s[110],l[110],r[110];
while(~scanf("%s",dt[t]))join(dt[t++]);
for(int i=0;i<t;i++){
int len=strlen(dt[i]);
if(len<=1)continue;
for(int j=1;j<len;j++){
strcpy(r,dt[i]+j);
strcpy(l,dt[i]);
l[j]='\0';
if(find(l)&&find(r)){
puts(dt[i]);break;
}
}
}
return 0;
}

#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;
}