However, this is not fast enough. The team is going to put together a dictionary of the common words that a user may type. The goal is to reduce the average number of keystrokes needed to type words that are in the dictionary. During the typing of a word, whenever the following letter is uniquely determined, the cellphone system will input it automatically, without the need for a keystroke. To be more precise, the behavior of the cellphone system will be determined by the following rules:
For instance, if the dictionary is composed of the words `hello‘, `hell‘, `heaven‘ and `goodbye‘, and the user presses `h‘, the system will input `e‘ automatically, because every word which starts with `h‘ also starts with `he‘. However, since there are words that start with `hel‘ and with `hea‘, the system now needs to wait for the user. If the user then presses `l‘, obtaining the partial word `hel‘, the system will input a second `l‘ automatically. When it has `hell‘ as input, the system cannot guess, because it is possible that the word is over, or it is also possible that the user may want to press `o‘ to get `hello‘. In this fashion, to type the word `hello‘ the user needs three keystrokes, `hell‘ requires two, and `heaven‘ also requires two, because when the current input is `hea‘ the system can automatically input the remainder of the word by repeatedly applying the second rule. Similarly, the word `goodbye‘ needs just one keystroke, because after pressing the initial `g‘ the system will automatically fill in the entire word. In this example, the average number of keystrokes needed to type a word in the dictionary is then (3 + 2 + 2 + 1)/4 = 2.00.
Your task is, given a dictionary, to calculate the average number of keystrokes needed to type a word in the dictionary with the new cellphone system.
N105). Each of the next N lines contains a non-empty string of at most 80 lowercase letters from the English alphabet, representing a word in the dictionary. Within each test case all words are different, and the sum of the lengths of all words is at most 106.
4 hello hell heaven goodbye 3 hi he h 7 structure structures ride riders stress solstice ridiculous
2.00 1.67 2.71
吐槽：有一年没写过字典树了，看以前的代码竟然感到很陌生，更可笑的是指针都不知道怎么用了。真是学得不如忘得快，但是好好看看，就很快懂了。
题目大意：给你n个不重复的字符串，每次敲一个字母，有相同前缀的字母都会自动出来，类似手机输入法。问你打出所有字符串平均需要敲多少下。如第一组样例。
hello 只需要敲3次，一次是h，一次是l，一次是o。 hell只需要敲2次，一次是h，一次是l。heaven需要敲2次，一次是h，一次是a。goodbye需要敲1次，即g。所以全部敲完是8次，平均值是2.00。
解题思路：用字典树模拟过程。我们在每个结点记录重复度rep，该结点下面有多少个分叉cross，是否是字符串结尾flag。对于分叉大于1的结点，我们需要加上该结点的重复度，表示需要在该结点下面的那些不同的字符处敲键盘。如果该结点同时是串尾的话，则需要减1，因为对于当前这个串，不需要在下面多敲一次就已经得到了。对于分叉不大于1的结点，我们判断它是否是串尾，如果是串尾，那么我们只需要加上重复度-1即可。最后加上串的个数n，表示对于每个字串，都需要敲1次最开始的那个字母。
#include<bits/stdc++.h>
using namespace std;
struct NODE {
int rep;
int cross;
bool flag;
NODE *next[26];
};
typedef NODE Trie;
NODE *newnode(){
Trie *tmp;
tmp= new Trie;
for(int i=0;i<26;i++)
tmp->next[i]=NULL;
tmp->flag=0;
tmp->cross=0;
tmp->rep=0;
return tmp;
}
void Insert ( Trie *rt, char *s ) {
int len = strlen ( s );
int idx;
for ( int i = 0; i < len; i++ ) {
idx = s[i] - ‘a‘;
if(rt->next[idx] == NULL)
{
rt->next[idx] = newnode() ;
rt->cross++;
}
rt=rt->next[idx];
rt->rep++;
}
rt->flag=1;
}
double dfs(Trie *rt){
double ret=0;
for(int i=0;i<26;i++){
if(rt->next[i]!=NULL){
ret+=dfs(rt->next[i]);
}
}
if(rt->cross>1){
ret+=rt->rep;
if(rt->flag==1){
ret--;
}
}else if(rt->flag==1){
ret+=rt->rep-1;
}
delete rt;
return ret;
}
int main() {
int n;
char str[100];
while ( scanf ( "%d", &n ) != EOF ) {
Trie *root;
root=newnode();
for ( int i = 0; i < n; i++ ) {
scanf ( "%s", str );
Insert ( root, str );
}
double ans=dfs(root);
printf("%.2lf\n",(n+ans)/n*1.0);
}
return 0;
}
BNU 27847——Cellphone Typing——————【字典树】
原文：http://www.cnblogs.com/chengsheng/p/4712558.html