题目链接：点击打开链接

题意：给定n个字符串，选尽可能多的字符串使得每种字母出现的次数为偶数次

思路：

中途相遇法

import java.io.PrintWriter;
import java.util.ArrayList;
import java.util.Arrays;
import java.util.Collections;
import java.util.Comparator;
import java.util.Iterator;
import java.util.LinkedList;
import java.util.PriorityQueue;
import java.util.Scanner;
import java.util.TreeMap;
import java.util.TreeSet;
import java.util.Queue;

public class Main {
	class Node implements Comparable<Node>{
		int state, num;
		Node(){}
		Node(int state, int num){
			this.state = state; this.num = num;
		}
		public int compareTo(Node o) {
			return Integer.compare(state, o.state);
		}
		public Node clone(){
			return new Node(state, num);
		}
	}
	static int N = 25;
	int n;
	int[] a = new int[N];
	TreeMap<Integer,Integer> b = new TreeMap();
	int One(int x){
		int ans = 0;
		while(x>0){
			ans++;
			x&=x-1;
		}
		return ans;
	}
	int ans;
	void input(){
		n = cin.nextInt();
		for(int i = 0; i < n; i++){
			String s = "1";
			while(s.charAt(0)<'A'||s.charAt(0)>'Z')s = cin.next();
			a[i] = 0;
			for(int j = 0; j < s.length(); j++)
				a[i]^=(1<<(s.charAt(j)-'A'));			
		}
		b.clear();
		for(int i = 0; i < (1<<(n/2)); i++)
		{
			int tmp = 0;
			for(int j = 0; j < (n/2); j++)
				if((i&(1<<j)) > 0)
					tmp ^= a[j];
			if(!b.containsKey(tmp) || One(i) > One(b.get(tmp)))
				b.put(tmp, i);			
		}
	}
	void work() {
		while (cin.hasNext()) {
			ans = 0;
			input();
			for(int i = 0; i < (1<<(n-n/2)); i++)
			{
				int tmp = 0;
				for(int j = 0; j < n-n/2; j++)
					if((i&(1<<j))>0)
						tmp ^= a[j+n/2];
				if(b.containsKey(tmp) && One(i) + One(b.get(tmp)) > One(ans))
					ans = (i<<(n/2)) ^ b.get(tmp);
			}
			out.println(One(ans));
			for(int j = 0; j < n; j++)
				if((ans&(1<<j))>0)
					out.print((j+1)+" ");
			out.println();
		}
	}

	Main() {
		cin = new Scanner(System.in);
		out = new PrintWriter(System.out);
	}

	public static void main(String[] args) {
		Main e = new Main();
		e.work();
		out.close();
	}

	public Scanner cin;
	public static PrintWriter out;
}

UVA 1326 Jurassic Remains 中途相遇法