题意不难理解，就是给出一些断言，以及一些查询，回答查询或者在找到断言矛盾以后沉默不做任何事。
这题其实就是一个并查集的距离存储问题，只要记录并查集元素的相对值以及绝对值就可以了。
代码如下：
1 #include <iostream> 2 #include <algorithm> 3 #include <cstdio> 4 #include <cstring> 5 #include <stack> 6 #include <set> 7 8 using namespace std; 9 10 const int N = 22222; 11 struct MFS { 12 int fa[N], rel[N], val[N]; 13 void init() { for (int i = 0; i < N; i++) fa[i] = i, rel[i] = 0, val[i] = -1;} 14 int find(int x) { 15 if (fa[x] != x) { 16 int fx = find(fa[x]); 17 rel[x] ^= rel[fa[x]]; 18 fa[x] = fx; 19 val[x] = val[fx] ^ rel[x]; 20 return fx; 21 } else return x; 22 } 23 bool merge(int x, int y, int d) { 24 int fx = find(x), fy = find(y); 25 if (fx == fy) return d == (rel[x] ^ rel[y]); 26 int dx = rel[x] ^ rel[fx], dy = rel[y] ^ rel[fy]; 27 if (~val[fx] && ~val[fy]) return d == (dx ^ dy ^ val[fx] ^ val[fy]); 28 d ^= dx ^ dy; 29 if (~val[fx]) rel[fy] ^= d, fa[fy] = fx; 30 else rel[fx] ^= d, fa[fx] = fy; 31 return true; 32 } 33 bool confirm(int x, int d) { 34 int fx = find(x); 35 if (~val[fx]) return (rel[x] ^ val[fx]) == d; 36 val[fx] = rel[x] ^ d; 37 return true; 38 } 39 int query(int x) { 40 int fx = find(x); 41 if (~val[x]) return val[x]; 42 return -1; 43 } 44 } mfs; 45 46 set<int> nsid; 47 int ns[N]; 48 49 int main() { 50 // freopen("in", "r", stdin); 51 char buf[222]; 52 int n, m, val[111], cnt, cas = 1; 53 while (~scanf("%d%d", &n, &m) && (n || m)) { 54 mfs.init(); 55 bool conf = false; 56 char *ptr; 57 cnt = 0; 58 printf("Case %d: ", cas++); 59 for (int q = 1; q <= m; q++) { 60 scanf("%s", buf); 61 gets(buf + 1); 62 val[0] = 0; 63 ptr = strtok(buf + 1, " "); 64 while (ptr) { 65 val[0]++; 66 sscanf(ptr, "%d", &val[val[0]]); 67 ptr = strtok(NULL, " "); 68 } 69 if (conf) continue; 70 if (buf[0] == 'I') { 71 cnt++; 72 if (val[0] == 3) conf = !mfs.merge(val[1], val[2], val[3]); 73 else conf = !mfs.confirm(val[1], val[2]); 74 if (conf) printf("The first %d facts are conflicting. ", cnt); 75 } else { 76 int ans = 0, tmp, fa; 77 nsid.clear(); 78 for (int i = 2; i <= val[0]; i++) { 79 tmp = mfs.query(val[i]); 80 fa = mfs.find(val[i]); 81 // cout << tmp << ' '; 82 if (tmp < 0) { 83 if (nsid.find(fa) != nsid.end()) { 84 ans ^= ns[fa] ^ mfs.rel[val[i]]; 85 nsid.erase(fa); 86 } else { 87 nsid.insert(fa); 88 ns[mfs.fa[val[i]]] = mfs.rel[val[i]]; 89 } 90 } else ans ^= tmp; 91 } 92 // cout << endl; 93 if (!nsid.empty()) puts("I don't know."); 94 else printf("%d ", ans); 95 } 96 } 97 puts(""); 98 } 99 return 0; 100 }
——written by Lyon