codeforces-771C-Bear and Tree Jumps【树上计数】

看别人代码学习的,想了好久才看懂代码,自己写了一遍,又写点注释。

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <vector>
 4 using namespace std;
 5 int N,k;
 6 long long ans  = 0;  // 答案会爆掉 long long
 7 int sum[222222],num[222222][6],in[222222];
 8 //sum子树节点个数 num[u][i] u子树中,距离根节点距离mod k == i 的节点个数
 9 vector<int> E[222222];
10 void dfs(int u,int depth) //深度用来求两片区域相隔的距离
11 {
12     sum[u] = num[u][depth%k] = 1;
13     for (unsigned int i = 0; i<E[u].size();i++)
14     {
15         int v = E[u][i];
16         if (in[v]==1) continue;
17         in[v] = 1;
18         dfs(v,depth+1);
19         for (int j = 0; j<k ; j++)
20             for (int kk = 0; kk<k; kk++)
21         {
22             int t = (k- (((  ( j + kk- (2*depth) ) % k ) + k ) % k )) % k;
23             ans+=t*(long long)num[u][j]*(long long)num[v][kk]; // 将走的过程中 多余(浪费)的边计数
24         }
25         for (int j = 0; j<k; j++)
26             num[u][j]+=num[v][j];
27         sum[u]+=sum[v];
28     }
29     ans+=(long long)sum[u]*(long long)(N-sum[u]); // 统计每条边需要走的次数
30 }
31 int main()
32 {
33     cin >> N >> k;
34     for (int i = 1; i<N; i++)
35     {
36         int a,b;
37         scanf("%d%d",&a,&b);
38         E[a].push_back(b);
39         E[b].push_back(a);
40     }
41     in[1] = 1;
42     dfs(1,0);
43     cout << ans/k <<endl;
44 }
原文地址:https://www.cnblogs.com/HITLJR/p/6590448.html