CodeForces - 771C Bear and Tree Jumps

题意：

给定一棵有 $n$ 个结点的树，边权均为 $1$，再给定 $k$，求 $\sum\limits_{i}^{n}\sum\limits_{j=i+1}^{n}\lceil\cfrac{dis(i, j)}{k}\rceil$。 $(n \leq 2 × 10^5, k \leq 5)$

链接：

https://codeforces.com/problemset/problem/771/C

解题思路：

树形 $dp$，将各路径依照长度模 $k$ 分开存储，合并求解时枚举 $k^2$ 种路径组合即可。

参考代码：

#include<bits/stdc++.h>
 
using namespace std;
typedef long long ll;
typedef pair<int, int> pii;
#define sz(a) ((int)a.size())
#define pb push_back
#define lson (rt << 1)
#define rson (rt << 1 | 1)
#define gmid (l + r >> 1)
const int maxn = 2e5 + 5;
const int inf = 0x3f3f3f3f;
const int mod = 1e9 + 7;
 
vector<int> G[maxn];
ll dp[maxn][5], ans;
int siz[maxn][5];
int n, k;

void dfs(int u, int f){

    siz[u][0] = 1;
    for(auto &v : G[u]){

        if(v == f) continue;
        dfs(v, u);
        for(int i = 0; i < k; ++i){

            siz[0][(i + 1) % k] = siz[v][i];
            dp[0][(i + 1) % k] = dp[v][i] + siz[v][i];
        }
        for(int i = 0; i < k; ++i){

            for(int j = 0; j < k; ++j){

                ll d1 = dp[u][i] - i * siz[u][i], d2 = dp[0][j] - j * siz[0][j];
                ans += (d1 * siz[0][j] + d2 * siz[u][i]) / k;
                int dt = (i + j + k - 1) / k;
                ans += dt * 1ll * siz[0][j] * siz[u][i];
            }
        }
        for(int i = 0; i < k; ++i){

            siz[u][i] += siz[0][i];
            dp[u][i] += dp[0][i];
        }
    }
}

int main(){
 
    ios::sync_with_stdio(0); cin.tie(0);
    cin >> n >> k;
    for(int i = 1; i < n; ++i){

        int u, v; cin >> u >> v;
        G[u].pb(v), G[v].pb(u);
    }
    dfs(1, 0);
    // for(int i = 1; i <= n; ++i){

    //     for(int j = 0; j < k; ++j){

    //         cout << i << " " << j << " " << siz[i][j] << " " << dp[i][j] << endl;
    //     }
    // }
    cout << ans << endl;
    return 0;
}