Exclusive-OR
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 2177 Accepted Submission(s): 603
Problem Description You are
not given
n non-negative integers
X0, X1, ..., Xn-1 less than 2
20 , but they do exist, and their values never change.
I'll gradually provide you some facts about them, and ask you some questions.
There are two kinds of facts, plus one kind of question:
Input There will be at most 10 test cases. Each case begins with two integers
n and
Q (1 <=
n <= 20,000, 2 <=
Q <= 40,000). Each of the following lines contains either a fact or a question, fZ??http://www.2cto.com/kf/ware/vc/" target="_blank" class="keylink">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"t deduce the answer for a particular question, from the facts I provide you
before that question, print "I don't know.", without quotes. If the
i-th fact (don't count questions)
cannot be consistent with
all the facts before that, print "The first i facts are conflicting.", then keep silence for everything after that (including facts and questions). Print a blank line after the output of each test case.
Sample Input
2 6
I 0 1 3
Q 1 0
Q 2 1 0
I 0 2
Q 1 1
Q 1 0
3 3
I 0 1 6
I 0 2 2
Q 2 1 2
2 4
I 0 1 7
Q 2 0 1
I 0 1 8
Q 2 0 1
0 0
Sample Output
Case 1:
I don't know.
3
1
2
Case 2:
4
Case 3:
7
The first 2 facts are conflicting.
Source 2009 Asia Wuhan Regional Contest Hosted by Wuhan University
题意：
有n(n<=20000)个未知的整数X0,X1,X2Xn-1，有以下Q个(Q<=40000)操作：
I p v :告诉你Xp=v
I p q v :告诉你Xp ^ Xq=v
Q k p1 p2 … pk : 询问 Xp1 Xor Xp2 .. Xor Xpk， k不大于15。
如果当前的I跟之前的有冲突的话，跳出
思路：
并查集，每个节点记录与根节点的异或偏移量，一个集合内如果有一个知道值了的话，这个集合里面都能知道值，（可以标记根是否已经得到值，或者像网上大部分人的做法，虚拟出一个节点n，值为0，将I操作统一）查询时，如果一个未知集合出现了偶数个，那么可以得到其值，u^root^v^root=u^v，如果出现奇数次，那么不能得到。Merge的时候有个小坑，见代码。
代码：
#include
#include
#include
#include
#include
#include
#include