Description

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 off(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 most2. 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).

There are  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.

题意:给出一棵树,和最多跳K个数字,f(s,t)表示从s到t需要跳的最少次数,问那么一棵树每两个点跳的次数之和是多少?

解法:如果只跳一次,那每个点经历的次数为这个点的子树节点个数*(n-这个点的子树节点个数),那么K>=2的情况,对于任意两个点的x->y 距离为 深度[x]+深度[y]-2*最近公共祖先深度[z]

dp[v][d%k]表示v为节点开始,深度为d%k的个数,比如2为节点开始,深度为1的有4和5

4在2的子树内,4开始深度为1点为6,我们更新到dp[2][1]内,就是dp[2][1]+=dp[4][1]

sum[v]表示v的子树节点个数

最后答案为ans/n,比如距离为5,最多跳3步,其实是跳(5+1)/3=2次就好了,注释也有解释

(题解好难懂啊,我也不知道说清楚了没)

 #include<bits/stdc++.h>
using namespace std;
#define ll long long
const int maxn=;
ll dp[maxn][],sum[maxn];
vector<int>q[maxn];
ll vis[maxn];
ll cnt;
ll n,k;
void dfs(int v,int d,int fa)
{
dp[v][d%k]=sum[v]=;
//当前子节点就v一个
for(int i=;i<q[v].size();i++)
{
// cout<<v<<endl;
int pos=q[v][i];
if(pos==fa) continue;
dfs(pos,d+,v);
for(int x=;x<k;x++)
{
for(int y=;y<k;y++)
{
int ans=((x+y)%k-(d*)%k+k)%k;
//ans为缺少部分,比如5跳3,少了两步
cnt+=((k-ans)%k)*dp[v][x]*dp[pos][y];
//少了两步,为了达成三步,必须多走一步,所以为k-ans,每个点多走k-ans步,相乘
}
}
for(int x=;x<k;x++)
{
dp[v][x]+=dp[pos][x];
//子节点记录部分更新到父结点
}
cnt+=sum[pos]*(n-sum[pos]);
//讨论k=1的情况,种数==为pos节点包含子节点*以外的节点
sum[v]+=sum[pos];
//将pos包含节点个数更新到父结点 }
}
int main()
{
cin>>n>>k;
for(int i=;i<n-;i++)
{
int u,v;
cin>>u>>v;
q[u].push_back(v);
q[v].push_back(u);
}
dfs(,,);
cout<<cnt/k<<endl;
return ;
}

Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) D的更多相关文章

  1. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 菜鸡只会ABC!

    Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) 全场题解 菜鸡只会A+B+C,呈上题解: A. Bear and ...

  2. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) C. Bear and Different Names 贪心

    C. Bear and Different Names 题目连接: http://codeforces.com/contest/791/problem/C Description In the arm ...

  3. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) B - Bear and Friendship Condition 水题

    B. Bear and Friendship Condition 题目连接: http://codeforces.com/contest/791/problem/B Description Bear ...

  4. 【树形dp】Codeforces Round #405 (rated, Div. 1, based on VK Cup 2017 Round 1) B. Bear and Tree Jumps

    我们要统计的答案是sigma([L/K]),L为路径的长度,中括号表示上取整. [L/K]化简一下就是(L+f(L,K))/K,f(L,K)表示长度为L的路径要想达到K的整数倍,还要加上多少. 于是, ...

  5. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1)

    A 模拟 B 发现对于每个连通块,只有为完全图才成立,然后就dfs C 构造 想了20分钟才会,一开始想偏了,以为要利用相邻NO YES的关系再枚举,其实不难.. 考虑对于顺序枚举每一个NO/YES, ...

  6. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1)A B C 水 并查集 思路

    A. Bear and Big Brother time limit per test 1 second memory limit per test 256 megabytes input stand ...

  7. 【构造】Codeforces Round #405 (rated, Div. 1, based on VK Cup 2017 Round 1) A. Bear and Different Names

    如果某个位置i是Y,直接直到i+m-1为止填上新的数字. 如果是N,直接把a[i+m-1]填和a[i]相同即可,这样不影响其他段的答案. 当然如果前面没有过Y的话,都填上0就行了. #include& ...

  8. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) E

    Description Bear Limak prepares problems for a programming competition. Of course, it would be unpro ...

  9. Codeforces Round #405 (rated, Div. 2, based on VK Cup 2017 Round 1) C

    Description In the army, it isn't easy to form a group of soldiers that will be effective on the bat ...

随机推荐

  1. ViewGroup如何分发事件

    dispatchTouchEvent事件派发显示隧道方式.再是冒泡方式隧道方式传递,直道某一个元素消耗此事件,由上至下逐层分发视图.冒泡方式传递,当某个视图消耗事件后其return boolean 是 ...

  2. 树莓派wiringPi经常使用的函数介绍

     1.void pinMode (int pin, int mode) ; 这个函数式设置pin脚的输入和输出模式以及PWM的输入和输出模式.在wiringPi中仅仅有 pin 1 (BCM_GP ...

  3. dp求顺序hdu1160

    题意是仅仅求一次的顺序.先依照速度从大到小排序,速度想等到按体重增序排列. 然后基本就变成了求已定顺序序列的最长递增序列递增,跟那个求一致最大序列和的基本一致. dp[i]里存储的是到当前i最大的递增 ...

  4. JAVA运行时异常及常见的5中RuntimeExecption

    最近在抽时间看面试题,很多面试题都提出了写出java常见的5个运行时异常.现在来总结一下, java运行时异常是可能在java虚拟机正常工作时抛出的异常. java提供了两种异常机制.一种是运行时异常 ...

  5. A nonrecursive list compacting algorithm

    A nonrecursive list compacting algorithm Each Erlang process has its own stack and heap which are al ...

  6. react源码分析

    ReactMount.render -> ReactMount._renderSubtreeIntoContainer -> ReactMount._renderNewRootCompon ...

  7. CSS中的那点事儿(一)--- CSS中的单位1

    单位主要用在规定元素的数值性css属性的,这里主要讨论应用的比较多的是width height  padding margin font-size,而单位中应用最广泛就是%.px.em 百分比 百分比 ...

  8. POJ 2739 Sum of Consecutive Prime Numbers( *【素数存表】+暴力枚举 )

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 19895 ...

  9. hdu-2680 Choose the best route(最短路)

    题目链接: Choose the best route Time Limit: 2000/1000 MS (Java/Others)     Memory Limit: 32768/32768 K ( ...

  10. I.MX6 新版u-boot分析

    /******************************************************************* * I.MX6 新版u-boot分析 * 说明: * 因为一些 ...