题意：训练指南P245
分析：首先这道是经典的并查集题目，利用异或的性质。异或性质：x ^ 0 = x -> a ^ a = 0 -> x ^ a ^ a = x，即一个数对某个数异或偶数次等于它本身。
第一种操作：p = v，设立一个超级根节点RT，rt[p] = RT, edge[p] = v，表示p的父亲节点是RT，到它的值就是p本身的值(p^0)。第二种p q v操作，分别找p和q的祖先，将祖先合并，那么edge[r1] = v1 ^ v2 ^ v3，这样保证下一次v1'仍然是p到祖先的异或值，这样p ^ q = p ^ root' ^ q ^ root' = v1'^v2’。第三种操作，如果pi的祖先是RT，那么v1就是p的值，否则v1 = p ^ rootp，这样需要同一个祖先的个数是偶数才能将祖先的影响取消。
#includeusing namespace std;const int N = 2e4 + 5;const int Q = 4e4 + 5;struct DSU { int rt[N], edge[N]; void clear(void) { memset (rt, -1, sizeof (rt)); } int Find(int x, int &val) { if (rt[x] == -1) { val = 0; return x; } int root = Find (rt[x], val); val ^= edge[x]; edge[x] = val; rt[x] = root; return root; } void Union(int x, int y, int val) { if (x > y) swap (x, y); rt[x] = y; edge[x] = val; }}dsu;int id[N], RT;void error(int fac) { printf ("The first %d facts are conflicting.\n", fac);}int main(void) { int n, m, cas = 0; while (scanf ("%d%d", &n, &m) == 2) { if (!n && !m) break; printf ("Case %d:\n", ++cas); RT = n; //super root dsu.clear (); bool err = false; char str[30]; int p, q, k, v1, v2, v3, r1, r2; int facts = 0; for (int i=1; i<=m; ++i) { scanf ("%s", &str); if (str[0] == 'I') { facts++; gets (str); if (err) continue; if (sscanf (str, "%d %d %d", &p, &q, &v3) == 2) { v3 = q; int r1 = dsu.Find (p, v1); if (r1 == RT) { //super tree if (v1 != v3) { error (facts); err = true; } } else { dsu.Union (r1, RT, v1 ^ v3); } } else { int r1 = dsu.Find (p, v1); int r2 = dsu.Find (q, v2); if (r1 == r2) { //in the same tree if ((v1 ^ v2) != v3) { error (facts); err = true; } } else dsu.Union (r1, r2, v1 ^ v2 ^ v3); //cat r1 and r2 } } else { scanf ("%d", &k); for (int j=1; j<=k; ++j) { scanf ("%d", &id[j]); } if (err) continue; int ans = 0; vector rs; for (int j=1; j<=k; ++j) { int r1 = dsu.Find (id[j], v1); ans ^= v1; if (r1 != RT) { rs.push_back (r1); } } if (rs.size () % 2 == 1) { printf ("I don't know.\n"); } else { sort (rs.begin (), rs.end ()); for (int j=0; j