字典樹 - A Poet Computer

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The ACM team is working on an AI project called (Eih Eye Three) that allows computers to write poems. One of the problems they stumbled upon is finding words with the same suffix. The ACM team constructed a dictionary of words, They are interested only in the longest common suffix, That is, a suffix common to three or more words in the dictionary… A suffix is any substring that starts from some arbitrary position in the string and reaches the end of the string. As the ACM team was also preparing for the ACM-TCPC2015 contest, they figured that the contestants can help in solving this problem. Your task is to write a program that finds a longest common suffix in a dictionary of words. An entry in the dictionary is a word of English letters only. Small letters are the same as capital letters. You can assume that there is exactly one unique solution for every test case.

Input

The first line of the input contains an integer T, the number of test cases. Each test case starts with a line containing one integer K, then K lines follow, each containing one string “Si” that represents an entry in the dictionary. 0 < T ≤ 50 |S_{i| ≤ 100 0 < K ≤ 1000}

Output

For each test case, print on the first line “Case c:” where ‘c’ is the test case number. On the second line you should print an integer denoting the length of the longest common suffix and another integer denoting how many words have the suffix appeared in.

Sample Input

Input

2
4
cocochannel
chrisschannel
MBCchannel
controlpanel
5
superman
batman
ironman
chrissbrown
MyCrown

Output

Case 1:
7 3
Case 2:
3 3


--------------------------------------------------------------我是分割線^_^----------------------------------------------------------


一道字典樹的題目，當時還是想用指標做出來的，可惜沒有時間了，不過賽後看了一下別人的代碼，覺得直接
開結構體數組類比指標也蠻好的，雖然浪費很多記憶體，不過速度很快，值得一學，之前寫過指標的字典樹，覺
得指來指去真是會讓人頭暈= =，所以學習一下這個結構體數組版本的字典樹，其實和指標版的差不多，稍微
改改就行了，主體基本一樣的，以後慢慢理解.......記住這個套路就行

#include<iostream>#include<algorithm>#include<cstdio>#include<cstring>#include<string>#include<cmath>#include<vector>#include<queue>#include<cctype>using namespace std;#define Int __int64#define INF 0x3f3f3f3fconst int MAXN = 111111;int triecnt;int root;struct Node {    int End;//代表以此結為的字串有多少個    int son[26];    int deep;//字串的第幾個字元}trie[MAXN];int new_tire() {    triecnt++;    trie[triecnt].End = 0;    for (int i = 0; i < 26; i++) {        trie[triecnt].son[i] = 0;    }    return triecnt;}void init_trie() {    triecnt = 0;    root = new_tire();}void insert_str(char str[]) {    int len = strlen(str);    int rt = root;    for (int i = len - 1; i >= 0; i--) {        int id = str[i] - ‘a‘;        if (trie[rt].son[id] == 0) {            trie[rt].son[id] = new_tire();            rt = trie[rt].son[id];            trie[rt].End++;            trie[rt].deep = len - i;        } else {            rt = trie[rt].son[id];            trie[rt].End++;            trie[rt].deep = len - i;        }    }}int main() {    //freopen("input.txt", "r", stdin);    int cas;    int sign;    while (scanf("%d", &cas) != EOF) {        sign = 1;        while (cas--) {            init_trie();            int k;            scanf("%d", &k);            while (k--) {                char str[105];                scanf("%s", str);                int len = strlen(str);                for (int i = 0; i < len; i++) str[i] = tolower(str[i]);                insert_str(str);            }            int ans1 = 0, ans2 = 0;            for (int i = 1; i <= triecnt; i++) {                if (trie[i].End >= 3 && trie[i].deep > ans2) {                    ans1 = trie[i].End;                    ans2 = trie[i].deep;                }            }            printf("Case %d:\n%d %d\n", sign++, ans2, ans1);        }    }    return 0;}

字典樹 - A Poet Computer