[2019 Shenyang D network game title] Dawn-K's water (dotted rule)

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Title means obtained from the tree at any point of the 3 remaining and take.

Hear the game meaning of the questions that this is a point of divide and conquer, but write more wrong, before you think it will pay T, if it does not then the T. Minor changes result teammates wrote a hair dp directly over Orz.

Maintenance after the game thought something was too brain-dead, thinking like Luo Gu board can maintain the title as violence, but in reality be stuck.

After the game the idea is to maintain the current distance from the center of gravity to the modulo 3 as well as the number and distance. Then, when statistics enumerate two points from taking residual value, then the statistical contribution.

Note downstream time to eliminate duplicate contribution.

Code ugly please forgive me QAQ

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll;
 4 const int maxn = 3e4 + 15;
 5 const int mod = 1e9 + 7;
 6 struct node {
 7     int s, e, w, next;
 8 }edge[maxn * 2];
 9 int head[maxn], len;
10 void add(int s, int e, int w) {
11     edge[len].e = e;
12     edge[len].w = w;
13     edge[len].next = head[s];
14     head[s] = len++;
15 }
16 int n, root, sum;
17 int vis[maxn], f[maxn], son[maxn];
18 ll ans[4], o[4], num[4];
19 void getroot(int x, int fa) {
20     son[x] = 1, f[x] = 0;
21     for (int i = head[x]; i != -1; i = edge[i].next) {
22         int y = edge[i].e;
23         if (y == fa || vis[y])continue;
24         getroot(y, x);
25         son[x] += son[y];
26         f[x] = max(f[x], son[y]);
27     }
28     f[x] = max(f[x], sum - son[x]);
29     if (f[x] < f[root])root = x;
30 }
31 void getd(int x, int dis, int fa) {
32     o[dis % 3]++;
33     num[dis % 3] += dis;
34     for (int i = head[x]; i != -1; i = edge[i].next) {
35         int y = edge[i].e;
36         if (y == fa || vis[y])continue;
37         getd(y, (dis + edge[i].w) % mod, x);
38     }
39 }
40 void cal(int x, int val, int add) {
41     getd(x, val, 0);
42     for (int i = 0; i < 3; i++)
43         for (int j = 0; j < 3; j++) {
44             ans[(i + j) % 3] = (ans[(i + j) % 3] + o[i] * num[j] * add % mod + mod) % mod;
45             ans[(i + j) % 3] = (ans[(i + j) % 3] + o[j] * num[i] * add % mod + mod) % mod;
46         }
47     for (int i = 0; i < 3; i++)o[i] = num[i] = 0;
48 
49 
50 }
51 void solve(int x) {
52     cal(x, 0, 1);
53     vis[x] = 1;
54     for (int i = head[x]; i != -1; i = edge[i].next) {
55         int y = edge[i].e;
56         if (vis[y])continue;
57         cal(y, edge[i].w, -1);
58         sum = son[y];
59         root = 0;
60         getroot(y, 0);
61         solve(root);
62     }
63 }
64 int main() {
65     while (scanf("%d", &n) != EOF) {
66         len = 0;
67         for (int i = 0; i <= n + 10; i++)
68             vis[i] = 0, head[i] = -1;
69         for (int i = 1; i < n; i++) {
70             int x, y, z;
71             scanf("%d%d%d", &x, &y, &z);
72             x++, y++;
73             add(x, y, z);
74             add(y, x, z);
75         }
76         for (int i = 0; i < 3; i++)
77             ans[i] = 0;
78         root = 0, f[0] = INT_MAX - 1;
79         sum = n;
80         getroot(1, 0);
81         solve(root);
82         for (int i = 0; i <= 2; i++) {
83             if (i == 2)
84                 printf("%lld\n", ans[i]);
85             else
86                 printf("%lld ", ans[i]);
87         }
88     }
89 }