容易想到树链剖分来维护
一条链上维护儿子中是1的个数为1的点的最长值和儿子是1的个数为2的点的最长值
于是每次修改的时候就二分查询会更新到哪里，再直接链修改就好了
单次查询复杂度$O(logn^2)$，单次修改复杂度为$O(logn)$
注意如果动态开点太多会导致MLE，最后解决办法是在每个线段树节点上增加了一个res变量表示返回值
1 /**************************************************************
2 Problem: 3553
3 User: rausen
4 Language: C++
5 Result: Accepted
6 Time:18880 ms
7 Memory:102672 kb
8 ****************************************************************/
9
10 #include <cstdio>
11
12 using namespace std;
13 const int N = 5e5 + 5;
14
15 inline int read();
16
17 int n;
18 int seq[N * 3];
19
20 struct tree_node {
21 int v, dep, sz;
22 int son[3], fa, top, w;
23 } tr[N * 3];
24
25 struct seg {
26 seg *ls, *rs, *res;
27 int tag, mx[2], sz;
28
29 #define Len (1 << 16)
30 inline void* operator new (size_t) {
31 static seg *mempool, *c;
32 if (c == mempool)
33 mempool = (c = new seg[Len]) + Len;
34 c -> ls = c -> rs = c -> res = NULL;
35 c -> tag = -1, c -> sz = 1, c -> mx[0] = c -> mx[1] = 0;
36 return c++;
37 }
38 #undef Len
39
40 inline void fill(int x) {
41 tag = x;
42 mx[0] = mx[1] = 0;
43 if (tag == 1) mx[0] = sz;
44 if (tag == 2) mx[1] = sz;
45 }
46 inline void push() {
47 if (tag != -1) {
48 if (ls) ls -> fill(tag);
49 if (rs) rs -> fill(tag);
50 tag = -1;
51 }
52 }
53 inline void update(seg *t1, seg *t2) {
54 sz = t1 -> sz + t2 -> sz;
55 mx[0] = t2 -> mx[0] + (t2 -> mx[0] == t2 -> sz ? t1 -> mx[0] : 0);
56 mx[1] = t2 -> mx[1] + (t2 -> mx[1] == t2 -> sz ? t1 -> mx[1] : 0);
57 }
58
59 #define mid (l + r >> 1)
60 void build(int l, int r) {
61 if (l == r) {
62 fill(tr[seq[l]].v);
63 return;
64 }
65 ls = new()seg, rs = new()seg, res = new()seg;
66 ls -> build(l, mid), rs -> build(mid + 1, r);
67 update(ls, rs);
68 }
69
70 void modify(int l, int r, int L, int R, int d) {
71 if (L <= l && r <= R) {
72 fill(d);
73 return;
74 }
75 push();
76 if (L <= mid) ls -> modify(l, mid, L, R, d);
77 if (mid < R) rs -> modify(mid + 1, r, L, R, d);
78 update(ls, rs);
79 }
80
81 int query(int l, int r, int pos) {
82 if (l == r) return tag;
83 push();
84 if (pos <= mid) return ls -> query(l, mid, pos);
85 else return rs -> query(mid + 1, r, pos);
86 }
87 seg* query(int l, int r, int L, int R) {
88 if (L <= l && r <= R) return this;
89 push();
90 if (R <= mid) return ls -> query(l, mid, L, R);
91 if (mid < L) return rs -> query(mid + 1, r, L, R);
92 res -> update(ls -> query(l, mid, L, R), rs -> query(mid + 1, r, L, R));
93 return res;
94 }
95 #undef mid
96 } *T;
97
98 inline void modify(int p, int tar, int c) {
99 while (tr[p].top != tr[tar].top)
100 T -> modify(1, n, tr[tr[p].top].w, tr[p].w, c), p = tr[tr[p].top].fa;
101 T -> modify(1, n, tr[tar].w, tr[p].w, c);
102 }
103
104 int change(int p) {
105 static int c, now, q;
106 static seg *tmp;
107 c = tr[p].v, now = tr[p].fa, tr[p].v = 1 - tr[p].v;
108 while (1) {
109 tmp = T -> query(1, n, tr[tr[now].top].w, tr[now].w);
110 if (tmp -> sz != tmp -> mx[c]) break;
111 now = tr[now].top;
112 if (!tr[now].fa || T -> query(1, n, tr[tr[now].fa].w) != c + 1) break;
113 now = tr[now].fa;
114 }
115 if (T -> query(1, n, tr[now].w) != c + 1) {
116 T -> modify(1, n, tr[now].w, tr[now].w, T -> query(1, n, tr[now].w) + (c ? -1 : 1));
117 return T -> query(1, n, 1) >= 2;
118 }
119 if (now == tr[now].top) q = now;
120 else q = seq[tr[now].w - T -> query(1, n, tr[tr[now].top].w, tr[now].w) -> mx[c] + 1];
121 modify(tr[p].fa, q, tr[p].v + 1);
122 if (tr[q].fa) T -> modify(1, n, tr[tr[q].fa].w, tr[tr[q].fa].w, T -> query(1, n, tr[tr[q].fa].w) + (c ? -1 : 1));
123 return T -> query(1, n, 1) >= 2;
124 }
125
126 void get_seq() {
127 static int i, j, q[N], tot_q, tot_d, p, S;
128 q[tot_q = 1] = 1, tr[1].fa = 0;
129 for (i = 1; i <= n; ++i)
130 for (j = 0; j < 3; ++j)
131 if (tr[q[i]].son[j] <= n)
132 tr[q[++tot_q] = tr[q[i]].son[j]].dep = tr[q[i]].dep + 1;
133 for (i = 1; i <= n; ++i) tr[i].top = i, tr[i].sz = 1;
134 for (i = n; i; --i) tr[tr[q[i]].fa].sz += tr[q[i]].sz;
135
136 tr[0].sz = 0, tr[1].w = 1;
137 for (i = 1; i <= n; ++i) {
138 p = q[i], tot_d = tr[p].w;
139 for (j = S = 0; j < 3; ++j)
140 if (tr[p].son[j] <= n && tr[tr[p].son[j]].sz > tr[S].sz)
141 S = tr[p].son[j];
142 if (S)
143 tr[S].w = tot_d + 1, tot_d += tr[S].sz, tr[S].top = tr[p].top;
144 for (j = 0; j < 3; ++j)
145 if (tr[p].son[j] <= n && tr[p].son[j] != S)
146 tr[tr[p].son[j]].w = tot_d + 1, tot_d += tr[tr[p].son[j]].sz;
147 }
148 for (i = n; i; --i) if (tr[q[i]].v >= 2) ++tr[tr[q[i]].fa].v;
149 for (i = 1; i <= n; ++i) seq[tr[i].w] = i;
150 }
151
152 int main() {
153 int i, j, Q;
154 n = read();
155 for (i = 1; i <= n; ++i)
156 for (j = 0; j < 3; ++j)
157 tr[tr[i].son[j] = read()].fa = i;
158 for (i = n + 1; i <= 3 * n + 1; ++i)
159 tr[tr[i].fa].v += (tr[i].v = read());
160 get_seq();
161 T = new()seg, T -> build(1, n);
162 Q = read();
163 while (Q--)
164 printf("%d\n", change(read()));
165 return 0;
166 }
167
168 inline int read() {
169 static int x;
170 static char ch;
171 x = 0, ch = getchar();
172 while (ch < ‘0‘ || ‘9‘ < ch)
173 ch = getchar();
174 while (‘0‘ <= ch && ch <= ‘9‘) {
175 x = x * 10 + ch - ‘0‘;
176 ch = getchar();
177 }
178 return x;
179 }
(p.s. 论编程能力弱的后果。。。写了一下午，调了一晚上QAQQQ)
时间： 2024-10-28 20:02:20