poj 1703 java

import java.io.*;public class test {static int a[];static int ans[];public static void main(String[] args)throws Exception {BufferedReader buf = new BufferedReader(new InputStreamReader(System.in));int test = Integer.parseInt(buf.readLine());for(int t=0;t<test;t++){String str[] = buf.readLine().split(" ");int n = Integer.parseInt(str[0]);int m = Integer.parseInt(str[1]);a = new int[n+1];ans = new int[n+1];for(int i=1;i<=n;i++)a[i] = i;for(int i=0;i<m;i++){str = buf.readLine().split(" ");int x = Integer.parseInt(str[1]);int y = Integer.parseInt(str[2]);if(str[0].equals("D")){Union(x,y);}else{if(n == 2)System.out.println("In different gangs.");else if(find(x)==find(y)){if(ans[x]==ans[y])System.out.println("In the same gang.");elseSystem.out.println("In different gangs.");}elseSystem.out.println("Not sure yet.");}}}}public static int find(int x) {if(a[x]==x)return x;int rt = find(a[x]);ans[x] = (ans[x]+ans[a[x]])%2;    // 类别偏移量         假设 之前 ans[fx]被赋值为1  因为原为0，所以在其下的子结点都需要改为与原来相反的值return a[x] = rt;}public static void Union(int x, int y) {int fx = find(x);int fy = find(y);a[fx] = fy;ans[fx] = (ans[x]+ans[y]+1)%2;        // 祖先结点 的ans值永远是 0     所以由x,y两点来判断 它们祖先结点 是否在同一部落                                      //ans[x],ans[y] 都为0或1 说明都跟自己的祖先结点 同一部落或不同部落  则 x的祖先与y的祖先也应该}                                         //在不同部落  此时 ans[fx] = 1;ans[fy]值 依旧是0}