HDU 3234

/* * 解法: 带权并查集神题. * 设置了一个超级根, 如果树中的一个数可以确认, 则将这棵树连接到超级根上. * 这样就简化了算法复杂性. */ #include <iostream> #include <algorithm> using namespace std; const int maxn = 20005; int N, Q, pare[maxn], cnt; __int64 edge[maxn]; bool flag; void init() { cnt = 0; flag = 1; memset(pare, -1, sizeof(pare)); } void Conflict() { flag = 0; printf("The first %d facts are conflicting./n", cnt); } void find(int x, int &r, __int64 &v) { if (pare[x] == -1) { r = x; v = 0; } else { find(pare[x], r, v); v = v ^ edge[x]; edge[x] = v; pare[x] = r; } } void mearge(int x, int y, __int64 v) { if (x > y) swap(x, y); pare[x] = y; edge[x] = v; } void Ifunc() { char str[20]; __int64 a, b, c; gets(str); if (!flag) return; cnt++; if (sscanf(str, "%I64d %I64d %I64d", &a, &b, &c) == 3) { int r1, r2; __int64 v1, v2; find(a, r1, v1); find(b, r2, v2); if (r1 == r2) { if (c != (v1 ^ v2)) { Conflict(); } } else { mearge(r1, r2, v1 ^ v2 ^ c); } } else { int r; __int64 v; find(a, r, v); if (r == N) { if (b != v) { Conflict(); } } else { mearge(r, N, v ^ b); } } } void Qfunc() { int k, x, i, r, ary[16], idx = 0; __int64 ans = 0, v; scanf("%d", &k); for (i = 0; i < k; i++) { scanf("%d", &x); find(x, r, v); ans = ans ^ v; if (r != N) ary[idx++] = r; } getchar(); if (!flag) return; if (idx % 2 == 1) { printf("I don't know./n"); return; } sort(ary, ary + idx); for (i = 0; i < idx; i += 2) if (ary[i] != ary[i+1]) break; if (i != idx) printf("I don't know./n"); else printf("%I64d/n", ans); } void InputCmd() { char cmd; scanf("%c", &cmd); if (cmd == 'I') Ifunc(); else Qfunc(); } int main() { int i, cas = 1; while (scanf("%d %d", &N, &Q), N || Q) { getchar(); printf("Case %d:/n", cas++); init(); for (i = 0; i < Q; i++) { InputCmd(); } printf("/n"); } return 0; }