首页 > 代码库 > BZOJ2819 Nim
BZOJ2819 Nim
做法。。。就不讲了,参见hzwer的blog好了
我们发现只要维护树上点到根的xor值就可以了,于是先搞个dfs序,然后用树状数组维护即可。
反正各种调不出。。。各种WA
后来发现又是LCA的姿势不对= =,今天不是刚写过noip题嘛T T
蒟蒻还是滚去挖矿算了、、、
1 /************************************************************** 2 Problem: 2819 3 User: rausen 4 Language: C++ 5 Result: Accepted 6 Time:10240 ms 7 Memory:76648 kb 8 ****************************************************************/ 9 10 #include <cstdio> 11 #include <algorithm> 12 13 #define lowbit(x) x & -x 14 using namespace std; 15 const int N = 500005; 16 17 struct edge { 18 int next, to; 19 edge() {} 20 edge(int _n, int _t) : next(_n), to(_t) {} 21 } e[N << 1]; 22 23 struct tree_node { 24 int v, dep, st, ed; 25 int fa[19]; 26 } tr[N]; 27 28 int n; 29 int tot, first[N]; 30 int cnt_seq, BIT[N]; 31 32 inline int read() { 33 int x = 0; 34 char ch = getchar(); 35 while (ch < ‘0‘ || ‘9‘ < ch) 36 ch = getchar(); 37 while (‘0‘ <= ch && ch <= ‘9‘) { 38 x = x * 10 + ch - ‘0‘; 39 ch = getchar(); 40 } 41 return x; 42 } 43 44 void XOR(int x, int v) { 45 while (x <= n) 46 BIT[x] ^= v, x += lowbit(x); 47 } 48 49 int query(int x) { 50 int res = 0; 51 while (x) 52 res ^= BIT[x], x -= lowbit(x); 53 return res; 54 } 55 56 inline void Add_Edges(int x, int y) { 57 e[++tot] = edge(first[x], y), first[x] = tot; 58 e[++tot] = edge(first[y], x), first[y] = tot; 59 } 60 61 void dfs(int p) { 62 int x, y; 63 for (x = 1; x <= 18; ++x) 64 tr[p].fa[x] = tr[tr[p].fa[x - 1]].fa[x - 1]; 65 tr[p].st = ++cnt_seq; 66 for (x = first[p]; x; x = e[x].next) 67 if ((y = e[x].to) != tr[p].fa[0]) { 68 tr[y].fa[0] = p, tr[y].dep = tr[p].dep + 1; 69 dfs(y); 70 } 71 tr[p].ed = cnt_seq; 72 XOR(tr[p].st, tr[p].v), XOR(tr[p].ed + 1, tr[p].v); 73 } 74 75 int lca(int x, int y) { 76 int i; 77 if (tr[x].dep < tr[y].dep) swap(x, y); 78 for (i = 18; ~i; --i) 79 if (tr[tr[x].fa[i]].dep >= tr[y].dep) 80 x = tr[x].fa[i]; 81 if (x == y) return x; 82 for (i = 18; ~i; --i) 83 if (tr[x].fa[i] != tr[y].fa[i]) 84 x = tr[x].fa[i], y = tr[y].fa[i]; 85 return tr[x].fa[0]; 86 } 87 88 int main() { 89 int i, x, y, LCA, Q; 90 char ch; 91 n = read(); 92 for (i = 1; i <= n; ++i) 93 tr[i].v = read(); 94 for (i = 1; i < n; ++i) 95 Add_Edges(read(), read()); 96 tr[1].dep = 1; 97 dfs(1); 98 Q = read(); 99 while (Q--) {100 ch = getchar();101 while (ch != ‘Q‘ && ch != ‘C‘) ch = getchar();102 x = read(), y = read();103 if (ch == ‘Q‘)104 puts(query(tr[x].st) ^ query(tr[y].st) ^ tr[lca(x, y)].v ? "Yes" : "No");105 else {106 XOR(tr[x].st, tr[x].v), XOR(tr[x].ed + 1, tr[x].v);107 tr[x].v = y;108 XOR(tr[x].st, tr[x].v), XOR(tr[x].ed + 1, tr[x].v);109 }110 }111 return 0;112 }
BZOJ2819 Nim
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。