HDU 3172 Virtual Friends (並查集)

Virtual Friends

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)

Total Submission(s): 2964 Accepted Submission(s): 861Problem DescriptionThese 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.

InputInput 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).

OutputWhenever 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

13Fred BarneyBarney BettyBetty Wilma

Sample Output

234

題意:問你就是到每句話的這兩個人的朋友網裡面有多少人

思路： 並查集

import java.util.*;import java.io.*;public class Main {int M = 100000;int n, m;HashMap<String, Integer> hm = new HashMap<String, Integer>();//用來給字串編號int patten[] = new int[M + 1];int ans[] = new int[M + 1];public static void main(String[] args) {new Main().work();}void work() {Scanner sc = new Scanner(new BufferedInputStream(System.in));while (sc.hasNext()) {hm.clear();n = sc.nextInt();for (int j = 0; j < n; j++) {//並查集初始化for (int i = 1; i <=M; i++) {patten[i] = i;}Arrays.fill(ans, 1);m = sc.nextInt();int count = 1;for (int i = 0; i < m; i++) {String s1 = sc.next();String s2 = sc.next();if (!hm.containsKey(s1)) {hm.put(s1, count);count++;}if (!hm.containsKey(s2)) {hm.put(s2, count);count++;}union(hm.get(s1), hm.get(s2));}}}}//並查集合并void union(int a, int b) {int pa = find(a);int pb = find(b);if (pa == pb) {System.out.println(ans[pa]);return;}if(pa>pb){patten[pb]=pa;ans[pa]+=ans[pb];System.out.println(ans[pa]);}else{patten[pa]=pb;ans[pb]+=ans[pa];System.out.println(ans[pb]);}}//並查集尋找int find(int x) {if(patten[x]==x)return x;int t=patten[x];patten[x]=find(t);//路徑最佳化return patten[x];}}