Let the Balloon Rise hdu1004 字典树

Problem Description

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Contest time again! How excited it is to see balloons floating around. But to tell you a secret, the judges’ favorite time is guessing the most popular problem. When the contest is over, they will count the balloons of each color and find the result.

This year, they decide to leave this lovely job to you.

Input

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Input contains multiple test cases. Each test case starts with a number N (0 < N <= 1000) – the total number of balloons distributed. The next N lines contain one color each. The color of a balloon is a string of up to 15 lower-case letters.

A test case with N = 0 terminates the input and this test case is not to be processed.

Output

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For each case, print the color of balloon for the most popular problem on a single line. It is guaranteed that there is a unique solution for each test case.

Author

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WU, Jiazhi

Source

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ZJCPC2004

Solution

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嗯，接触的第一道字典树题目
gdkoi被虐以后想了很多，看来还是努力比较实在，就从头开始了

字典树顾名思义就是一棵树，它以字母也不仅仅以字母作为节点，相比哈希能节省较多的空间，然后查找的复杂度只与查询字符串的长度有关

题目要求找出现最多的单词，那么开一棵trie然后加单词就行了，加完所有单词遍历一下。其实也是可以插入的同时查询最大值的，跑得贼快

Code

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#include <iostream>#include <string.h>#define rep(i, st, ed) for (int i = st; i <= ed; i += 1)#define fill(x, t) memset(x, t, sizeof(x))#define N 5001#define L 131using namespace std;int map[N][L], p[N], cnt = 0, ans;string prt;inline void insert(int now, string s, int dep){    if (dep == s.length()){        p[now] += 1;        if (p[now] > ans){            ans = p[now];            prt = s;        }        return;    }    if (!map[now][s[dep]]){        map[now][s[dep]] = ++ cnt;    }    insert(map[now][s[dep]], s, dep + 1);}int main(void){    ios::sync_with_stdio(false);    int n;    while (cin >> n && n){        ans = 0;        cnt = 0;        fill(map, 0);        fill(p, 0);        rep(i, 1, n){            string s;            cin >> s;            insert(0, s, 0);        }        cout << prt << endl;    }    return 0;}