Cellphone Typing
64-bit integer IO format: %lld Java class name: Main
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:
- The system never guesses the first letter of a word, so the first letter always has to be input manually by pressing the corresponding key.
- If a non-empty succession of letters c1c2...cn has been input, and there is a letter c such that every word in the dictionary which starts with c1c2...cn also starts with c1c2...cnc, then the system inputs c automatically, without the need of a keystroke. Otherwise, the system waits for the user.
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.
Input
Each test case is described using several lines. The first line contains an integer N representing the number of words in the dictionary ( 1
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.
Output
For each test case output a line with a rational number representing the average number of keystrokes needed to type a word in the dictionary. The result must be output as a rational number with exactly two digits after the decimal point, rounded if necessary.Sample Input
4 hello hell heaven goodbye 3 hi he h 7 structure structures ride riders stress solstice ridiculous
Sample Output
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+ans)/n*1.0);
}
return 0;
}