art != algo 傳參數 nag position des 下標 smallest

Description

World-renowned Prof. A. N. Agram‘s current research deals with large anagram groups. He has just found a new application for his theory on the distribution of characters in English language texts. Given such a text, you are to find the largest anagram groups.

A text is a sequence of words. A word w is an anagram of a word v if and only if there is some permutation p of character positions that takes w to v. Then, w and v are in the same anagram group. The size of an anagram group is the number of words in that group. Find the 5 largest anagram groups.

Input

The input contains words composed of lowercase alphabetic characters, separated by whitespace(or new line). It is terminated by EOF. You can assume there will be no more than 30000 words.

Output

Output the 5 largest anagram groups. If there are less than 5 groups, output them all. Sort the groups by decreasing size. Break ties lexicographically by the lexicographical smallest element. For each group output, print its size and its member words. Sort the member words lexicographically and print equal words only once.

Sample Input

undisplayed
trace
tea
singleton
eta
eat
displayed
crate
cater
carte
caret
beta
beat
bate
ate
abet

Sample Output

Group of size 5: caret carte cater crate trace .
Group of size 4: abet bate beat beta .
Group of size 4: ate eat eta tea .
Group of size 1: displayed .
Group of size 1: singleton .

Source

Ulm Local 2000 把每一個字符串，升序排序，可以得到一個字符串，如果兩個字符串的字典序最小的字符串相同，就屬於一個組， 所以用字典樹記錄這個最小字典序的字符串，然後映射下標到group結構體（存各種字符串，及個數，由於字符串輸出要去重，所以用set，這樣函數傳參數要引用傳遞，否則很慢），最後按照個數排序。 代碼：

#include <iostream>
#include <algorithm>
#include <cstdio>
#include <cstring>
#include <set>
#define MAX 30001
using namespace std;
struct group {
    int num;
    set<string> gr;
    group() {
        num = 0;
    }
}g[MAX];
int trie[MAX * 100][26],to[MAX * 100];
int pos,num;
int snum[26];
bool cmp(const group &a,const group &b) {///註意這裏
    if(a.num == b.num) return *(a.gr.begin()) < *(b.gr.begin());
    return a.num > b.num;
}
void Insert(char *s) {
    int len = strlen(s);
    string s1 = s;
    string s2 = "";
    for(int i = 0;s[i];i ++) {
        snum[s[i] - ‘a‘] ++;
    }
    for(int i = 0;i < 26;i ++) {
        while(snum[i]) {
            s2 += ‘a‘ + i;
            snum[i] --;
        }
    }
    int i = 0,c = 0;
    while(i < len) {
        int d = s2[i] - ‘a‘;
        if(!trie[c][d]) trie[c][d] = ++ pos;
        c = trie[c][d];
        i ++;
    }
    if(!to[c]) to[c] = ++ num;
    g[to[c]].gr.insert(s1) ;
    g[to[c]].num ++;
}

int main() {
    char str[100];
    while(~scanf("%s",str)) {
        Insert(str);
    }
    sort(g + 1,g + 1 + num,cmp);
    for(int i = 1;i <= 5;i ++) {
        printf("Group of size %d:",g[i].num);
        for(set<string>::iterator it = g[i].gr.begin();it != g[i].gr.end();it ++) {
            printf(" %s",(*it).c_str());
        }
        puts(" .");
    }
}

poj 2408 Anagram Groups