HDU 3460 Ancient Printer(java版)

题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=3460

题型:字典树
思路:
答案 ans=nodesum*2-max+n
nodesum:字典树中除根节点外的节点总数
max:最长单词长度
n:单词个数

ps:不知道为啥用Scanner就过不了,io就过了,也没有tle)玄学

import java.io.BufferedReader;
import java.io.IOException;
import java.io.InputStreamReader;
import java.io.OutputStreamWriter;
import java.io.PrintWriter;
import java.io.StreamTokenizer;

public class Main {
    static StreamTokenizer in = new StreamTokenizer(new BufferedReader(new InputStreamReader(System.in)));
    static PrintWriter out = new PrintWriter(new OutputStreamWriter(System.out));

    public static int nextInt() throws IOException {
        in.nextToken();
        return (int) in.nval;
    }

    public static String next() throws IOException {
        in.nextToken();
        return (String) in.sval;
    }

    public static void main(String[] args) throws IOException {

        while (in.nextToken()!=StreamTokenizer.TT_EOF) {
            int n = (int)in.nval;
            Trie trie = new Trie();
            int max = 0;
            for (int i = 0; i < n; i++) {
                String str =next();
                if (str.length() > max) {
                    max = str.length();
                }
                trie.insert(str);
            }
            System.out.println(trie.sum * 2 - max + n);
        }

    }
}

class Trie {
    private int size = 26;
    private TrieNode root;
     int sum;

    public Trie() {
        root = new TrieNode();
        sum=0;
    }

    private class TrieNode {
        private int num;
        private TrieNode[] son;
        private boolean isEnd;
        private char val;

        public TrieNode() {
            num = 1;
            son = new TrieNode[size];
            isEnd = false;
        }
    }

    public void insert(String str) {
        if (str == null || str.length() == 0)
            return;
        TrieNode node = root;
        char letters[] = str.toCharArray();
        for (int i = 0, len = str.length(); i < len; i++) {
            int pos = letters[i] - 'a';
            if (node.son[pos] == null) {
                node.son[pos] = new TrieNode();
                node.son[pos].val = letters[i];
                this.sum++;
            } else {
                node.son[pos].num++;
            }
            node = node.son[pos];
        }
        node.isEnd = true;
    }
    public TrieNode getRoot() {
        return this.root;
    }
}
HDOJ 1443是一个经典的数学模拟题目,通常涉及到动态规划。它描述的是“石头剪刀布”游戏的变种,玩家可以选择石头、剪刀或布,对手随机出拳。你需要编写一个程序来计算在给定的游戏轮数下,玩家获胜的概率。 以下是一个简化Java代码示例,实现了基本的策略并计算概率。假设我们使用朴素的策略,即每次都选择随机动作: ```java import java.util.Random; public class HDU_1443 { public static final String[] moves = {"石", "剪", "布"}; public static double getWinProbability(int n, int playerWins, int opponentWins) { Random random = new Random(); int totalGames = n * 3; double winCount = 0; for (int i = 0; i < totalGames; i++) { int playerMove = random.nextInt(3); int opponentMove = random.nextInt(3); if ((playerMove == 0 && opponentMove == 1) || (playerMove == 1 && opponentMove == 2) || (playerMove == 2 && opponentMove == 0)) { winCount++; } } return (double) winCount / totalGames * (1 + playerWins - opponentWins); // 考虑初始胜率差 } public static void main(String[] args) { int rounds = 1000; // 游戏轮数 System.out.println("Player wins on average in " + rounds + " games: " + getWinProbability(rounds, 0, 0)); // 初始胜率为0,两者无优势 } } ``` 注意:这是一个非常简单直接的解决方案,并未包含所有可能的策略优化,比如针对对手可能采取的特定策略进行调整。对于HDOJ这样的比赛题,可能需要更高级的策略分析。
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