Codeforces Round #405 Div. 1 B. Bear and Tree Jumps
2017-03-20 11:39
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A tree is an undirected connected graph without cycles. The distance between two vertices is the number of edges in a simple path between them.
Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.
Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.
For a pair of vertices (s, t) we define f(s, t) as
the minimum number of jumps Limak needs to get from s to t.
Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such
that s < t.
Input
The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) —
the number of vertices in the tree and the maximum allowed jump distance respectively.
The next n - 1 lines describe edges in the tree. The i-th
of those lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) —
the indices on vertices connected with i-th edge.
It's guaranteed that the given edges form a tree.
Output
Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such
that s < t.
Examples
input
output
input
output
input
output
Note
In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at
most 2. For example, from the vertex 5 he
can jump to any of vertices: 1, 2 and 4 (well,
he can also jump to the vertex 5 itself).
![](http://codeforces.com/predownloaded/9f/43/9f4358fb14a87a3173398425acc45177462691a8.png)
There are
![](http://codeforces.com/predownloaded/8c/85/8c853bfcd2a10642e5c24fea21bf99142f851491.png)
pairs
of vertices (s, t) such that s < t.
For 5 of those pairs Limak would need two jumps:(1, 6), (3, 4), (3, 5), (3, 6), (5, 6).
For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.
In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t),
so the answer is 3·1 = 3.
分析:一棵子树的所有路径可以分为过根和不过根节点,这题特殊在一步可以走k(k <= 5)条边,所以我们把所有路径按照模k划分为k类然后每次dfs时讨论一下就可以了,复杂度O(n*k^2).
Limak is a little polar bear. He lives in a tree that consists of n vertices, numbered 1 through n.
Limak recently learned how to jump. He can jump from a vertex to any vertex within distance at most k.
For a pair of vertices (s, t) we define f(s, t) as
the minimum number of jumps Limak needs to get from s to t.
Your task is to find the sum of f(s, t) over all pairs of vertices (s, t) such
that s < t.
Input
The first line of the input contains two integers n and k (2 ≤ n ≤ 200 000, 1 ≤ k ≤ 5) —
the number of vertices in the tree and the maximum allowed jump distance respectively.
The next n - 1 lines describe edges in the tree. The i-th
of those lines contains two integers ai and bi (1 ≤ ai, bi ≤ n) —
the indices on vertices connected with i-th edge.
It's guaranteed that the given edges form a tree.
Output
Print one integer, denoting the sum of f(s, t) over all pairs of vertices (s, t) such
that s < t.
Examples
input
6 2 1 2 1 3 2 4 2 5 4 6
output
20
input
13 3 1 2 3 2 4 2 5 2 3 6 10 6 6 7 6 13 5 8 5 9 9 11 11 12
output
114
input
3 5 2 1 3 1
output
3
Note
In the first sample, the given tree has 6 vertices and it's displayed on the drawing below. Limak can jump to any vertex within distance at
most 2. For example, from the vertex 5 he
can jump to any of vertices: 1, 2 and 4 (well,
he can also jump to the vertex 5 itself).
![](http://codeforces.com/predownloaded/9f/43/9f4358fb14a87a3173398425acc45177462691a8.png)
There are
![](http://codeforces.com/predownloaded/8c/85/8c853bfcd2a10642e5c24fea21bf99142f851491.png)
pairs
of vertices (s, t) such that s < t.
For 5 of those pairs Limak would need two jumps:(1, 6), (3, 4), (3, 5), (3, 6), (5, 6).
For other 10 pairs one jump is enough. So, the answer is 5·2 + 10·1 = 20.
In the third sample, Limak can jump between every two vertices directly. There are 3 pairs of vertices (s < t),
so the answer is 3·1 = 3.
分析:一棵子树的所有路径可以分为过根和不过根节点,这题特殊在一步可以走k(k <= 5)条边,所以我们把所有路径按照模k划分为k类然后每次dfs时讨论一下就可以了,复杂度O(n*k^2).
#include <bits/stdc++.h> #define MOD 10000007 #define N 200005 using namespace std; int n,k,u,v; long long ans,tf[5],tg[5],f [5],g [5]; vector<int> G ; void dfs(int u,int fa) { g[u][0]++; for(int i = 0;i < G[u].size();i++) { int v = G[u][i]; if(v != fa) { dfs(v,u); for(int j = 0;j < k;j++) { tf[(j+1) % k] = f[v][j] + ((j+1)/k)*g[v][j]; tg[(j+1) % k] = g[v][j]; } for(int j = 0;j < k;j++) for(int j2 = 0;j2 < k;j2++) ans += tf[j]*g[u][j2]+tg[j]*f[u][j2]+((bool)((j+j2)%k) + (j+j2)/k)*g[u][j2]*tg[j]; for(int j = 0;j < k;j++) { f[u][j] += tf[j]; g[u][j] += tg[j]; } } } } int main() { scanf("%d%d",&n,&k); for(int i = 1;i < n;i++) { scanf("%d%d",&u,&v); G[u].push_back(v); G[v].push_back(u); } dfs(1,1); cout<<ans<<endl; }
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