hdu3234 Exclusive-OR

Description

You are not given (n) non-negative integers (X_0, X_1, cdots, X_{n-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 le n le 20000, 2 le Q le 40000)) . Each of the following lines contains either a fact or a question, formatted as stated above. The (k) parameter in the questions will be a positive integer not greater than (15) , and the (v) parameter in the facts will be a non-negative integer less than (2^{20}). The last case is followed by (n=Q=0) , which should not be processed.

Output

For each test case, print the case number on its own line, then the answers, one on each one. If you can'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

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.

Solution

沉迷水题。

带权并查集。

#include<bits/stdc++.h>
using namespace std;

#define N 20001
#define rep(i, a, b) for (int i = a; i <= b; i++)

inline int read() {
	int x = 0, flag = 1; char ch = getchar(); while (!isdigit(ch)) { if (!(ch ^ '-')) flag = -1; ch = getchar(); }
	while (isdigit(ch)) x = (x << 1) + (x << 3) + ch - '0', ch = getchar(); return x * flag;
}

int n, Q, fa[N], num[N], val[N];

int find(int x) {
	if(x == fa[x]) return x;
	int t = fa[x]; fa[x] = find(fa[x]), val[x] ^= val[t]; return fa[x];
}

bool merge(int u, int v, int x) {
	int fu = find(u), fv = find(v); if(fu == fv) return (val[u] ^ val[v]) == x;
	if(fu == n) swap(fu, fv); fa[fu] = fv, val[fu] = val[u] ^ val[v] ^ x; return 1;
}

int main() {
	for (int T = 1; ; T++) {
		n = read(), Q = read(); if (!n && !Q) break; printf("Case %d:
", T);
		rep(i, 0, n) fa[i] = i, val[i] = 0;
		int facts = 0; bool flag = 1; char op[2], str[20];
		while (Q--) {
			if (!flag) { gets(str); continue; }
			scanf("%s", op);
			if(op[0] == 'I') {
				gets(str), facts++; int u, v, x;
				if(sscanf(str, "%d%d%d", &u, &v, &x) == 2) x = v, v = n;
				if(!merge(u, v, x)) printf("The first %d facts are conflicting.
", facts), flag = 0;
			}
			else {
				int k = read(), x[20], ans = 0; bool tag = 1;
				rep(i, 1, k) x[i] = read(), num[find(x[i])] = 0;
				rep(i, 1, k) num[find(x[i])]++, ans ^= val[x[i]];
				rep(i, 1, k) if(num[find(x[i])] % 2 == 1 && find(x[i]) != n) { tag = 0; break; }
				if(!tag) puts("I don't know."); else printf("%d
", ans);
			}
		}
	}
	return 0;
}