hdu 3172

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 7447    Accepted Submission(s): 2134

Problem Description

These days, you can do all sorts of things online. For example, you can use various websites to make virtual friends. For some people, growing their social network (their friends, their friends' friends, their friends' friends' friends, and so on), has become an addictive hobby. Just as some people collect stamps, other people collect virtual friends.

Your task is to observe the interactions on such a website and keep track of the size of each person's network.

Assume that every friendship is mutual. If Fred is Barney's friend, then Barney is also Fred's friend.

 

Input

Input file contains multiple test cases.
The first line of each case indicates the number of test friendship nest.
each friendship nest begins with a line containing an integer F, the number of friendships formed in this frindship nest, which is no more than 100 000. Each of the following F lines contains the names of two people who have just become friends, separated by a space. A name is a string of 1 to 20 letters (uppercase or lowercase).

 

Output

Whenever a friendship is formed, print a line containing one integer, the number of people in the social network of the two people who have just become friends.

 

Sample Input

1 3 Fred Barney Barney Betty Betty Wilma

 

Sample Output

2 3 4

 

Source

University of Waterloo Local Contest 2008.09

 

 

坑死我了。。开始以为是代码有问题，不停地WA，后来去参考别人的才发现是处理到文件尾。。。

while(sc.hasNext){  //千万要写这一层

　　int tcase = sc.nextInt();

　　while(tcase-->0){

　　　　int n = sc.nextInt();

　　　　while(n-->0){

　　　　}

　　}

}

 

思路就是把字符串映射+并查集.

我用了 TreeMap ，超时了,所以用了字典树

//A与B是朋友 ，B与C是朋友 则 A和C是朋友 ，所以 A,B的朋友圈里面有三个人，此题就是问在A，B的朋友圈里面有多少人
//题解： 字典树 +并查集

import java.util.Scanner;

class Trie_3127 {
    int key = 1;
    private Node root;

    public Trie_3127() {
        root = new Node();
    }

    int insert(String str) {
        Node t = root;
        int idx;
        for (int i = 0; i < str.length(); i++) {
            if (str.charAt(i) >= 'a' && str.charAt(i) <= 'z')
                idx = str.charAt(i) - 'a';
            else
                idx = str.charAt(i) - 'A' + 26;
            if (t.nodes[idx] == null) {
                Node node = new Node();
                t.nodes[idx] = node;
            }
            t = t.nodes[idx];
        }
        if (t.id == 0)
            t.id = key++;
        return t.id;
    }

    class Node {
        Node[] nodes;
        int id;

        public Node() {
            id = 0;
            nodes = new Node[52];
        }
    }
}

public class Main {
    static int[] father;
    static int[] friend;

    public static void main(String[] args) {
        Scanner sc = new Scanner(System.in);
        while (sc.hasNext()) {
            int tcase = sc.nextInt();
            while (tcase-- > 0) {
                Trie_3127 tree = new Trie_3127();
                int n = sc.nextInt();
                father = new int[2 * n + 1];
                friend = new int[2 * n + 1];
                for (int i = 1; i < 2 * n + 1; i++) {
                    father[i] = i; // i的根节点就是自己
                    friend[i] = 1; // i的朋友最初只有他自己一个
                }
                while (n-- > 0) {
                    String name1, name2;
                    name1 = sc.next();
                    name2 = sc.next();
                    int a = tree.insert(name1);
                    int b = tree.insert(name2);
                    // System.out.println(a+" "+b);
                    int x = find(a);
                    int y = find(b);
                    if (x != y) {
                        father[x] = y;
                        friend[y] += friend[x];
                    }
                    System.out.println(friend[y]);
                }
            }
        }

    }

    private static int find(int a) {
        if (a == father[a])
            return a;
        return find(father[a]);
    }
}