题意：树上进行两种操作：
（1）链上+d
（2）子树求和
直接链剖即可，子树求和就dfs序即可。
注意，dfs序按照重儿子在最先的序列，然后建立线段树。
1 /**************************************************************
2 Problem: 2836
3 User: rausen
4 Language: C++
5 Result: Accepted
6 Time:4180 ms
7 Memory:20264 kb
8 ****************************************************************/
9
10 #include <cstdio>
11 #include <algorithm>
12
13 using namespace std;
14 typedef long long ll;
15
16 const int N = 100005;
17
18 struct edge {
19 int next, to;
20 edge() {}
21 edge(int _n, int _t) : next(_n), to(_t) {}
22 } e[N];
23
24 struct tree_node {
25 int fa, son, sz, dep, st, ed, top;
26 } tr[N];
27
28 struct seg_node {
29 seg_node *ls, *rs;
30 int len;
31 ll v, tag;
32 } *root, mempool[N << 2], *cnt_seg = mempool, *null;
33
34 int n, Q;
35 int first[N], tot, cnt_q;
36
37 int read() {
38 int x = 0;
39 char ch = getchar();
40 while (ch < ‘0‘ || ‘9‘ < ch)
41 ch = getchar();
42 while (‘0‘ <= ch && ch <= ‘9‘)
43 (x *= 10) += ch - ‘0‘, ch = getchar();
44 return x;
45 }
46
47 inline void add_edge(int x, int y) {
48 e[++tot] = edge(first[x], y);
49 first[x] = tot;
50 }
51
52 #define y e[x].to
53 #define Son tr[p].son
54 void dfs(int p) {
55 int x;
56 tr[p].sz = 1;
57 for (x = first[p]; x; x = e[x].next) {
58 tr[y].dep = tr[p].dep + 1;
59 dfs(y);
60 tr[p].sz += tr[y].sz;
61 if (Son == 0 || tr[y].sz > tr[Son].sz)
62 Son = y;
63 }
64 }
65
66 void DFS(int p) {
67 int x;
68 tr[p].st = ++cnt_q;
69 if (Son) {
70 tr[Son].top = tr[p].top;
71 DFS(Son);
72 }
73 for (x = first[p]; x; x = e[x].next)
74 if (y != Son) {
75 tr[y].top = y;
76 DFS(y);
77 }
78 tr[p].ed = cnt_q;
79 }
80 #undef y
81 #undef Son
82
83 #define mid (l + r >> 1)
84 #define Ls p -> ls
85 #define Rs p -> rs
86 #define V p -> v
87 #define Tag p -> tag
88 #define Len p -> len
89 inline void push_tag(seg_node *p) {
90 Ls -> tag += Tag, Rs -> tag += Tag;
91 Ls -> v += 1ll * Tag * Ls -> len, Rs -> v += 1ll * Tag * Rs -> len;
92 Tag = 0;
93 }
94
95 inline void update(seg_node *p) {
96 V = Ls -> v + Rs -> v;
97 }
98
99 void seg_build(seg_node *&p, int l, int r) {
100 p = ++cnt_seg, Ls = Rs = null;
101 Len = r - l + 1;
102 V = Tag = 0;
103 if (l == r) return;
104 seg_build(Ls, l, mid), seg_build(Rs, mid + 1, r);
105 }
106
107 void seg_modify(seg_node *p, int l, int r, int L, int R, int d) {
108 if (l == L && R == r) {
109 Tag += d, V += 1ll * Len * d;
110 return;
111 }
112 push_tag(p);
113 if (L <= mid) seg_modify(Ls, l, mid, L, min(mid, R), d);
114 if (mid < R) seg_modify(Rs, mid + 1, r, max(mid + 1, L), R, d);
115 update(p);
116 }
117
118 ll seg_query(seg_node *p, int l, int r, int L, int R) {
119 if (l == L && R == r)
120 return V;
121 push_tag(p);
122 ll res = 0;
123 if (L <= mid) res += seg_query(Ls, l, mid, L, min(mid, R));
124 if (mid < R) res += seg_query(Rs, mid + 1, r, max(mid + 1, L), R);
125 return res;
126 }
127 #undef mid
128 #undef Ls
129 #undef Rs
130 #undef V
131 #undef Tag
132 #undef Len
133
134 void work_add(int x, int y, int d) {
135 while (tr[x].top != tr[y].top) {
136 if (tr[tr[x].top].dep < tr[tr[y].top].dep) swap(x, y);
137 seg_modify(root, 1, n, tr[tr[x].top].st, tr[x].st, d);
138 x = tr[tr[x].top].fa;
139 }
140 if (tr[x].dep < tr[y].dep) swap(x, y);
141 seg_modify(root, 1, n, tr[y].st, tr[x].st, d);
142 }
143
144 int main() {
145 int i, x, y, d, oper;
146 null = cnt_seg;
147 null -> ls = null -> rs = null;
148 n = read();
149 for (i = 1; i < n; ++i) {
150 x = read() + 1, y = read() + 1;
151 tr[y].fa = x;
152 add_edge(x, y);
153 }
154 dfs(1);
155 tr[1].top = 1;
156 DFS(1);
157 seg_build(root, 1, n);
158 Q = read();
159 while (Q--) {
160 oper = getchar();
161 while (oper != ‘A‘ && oper != ‘Q‘) oper = getchar();
162 if (oper == ‘A‘) {
163 x = read() + 1, y = read() + 1, d = read();
164 work_add(x, y, d);
165 }
166 else {
167 x = read() + 1;
168 printf("%lld\n", seg_query(root, 1, n, tr[x].st, tr[x].ed));
169 }
170 }
171 return 0;
172 }
时间： 2024-10-12 16:17:58