http://acm.hdu.edu.cn/showproblem.php?pid=5148
Problem Description
Long long ago,there is a knight called JayYe.He lives in a small country.This country is made up of n cities connected by n-1 roads(that means it‘s a tree).The king wants to reward JayYe because he beats the devil and save the princess.The king decide to give
JayYe exactly K cities as his daughter‘s dowry.
Here comes the question.Although JayYe beats the devil,his knee was injured.So he doesn‘t want to move too much,he wants his citys as close as possible,that is, JayYe wants the expected distance of his cities as small as possible.
The expected distance is defined as the expected distance between node u and node v,both u and v are randomly choose from JayYe‘s K cities equiprobably(that means first choose u randomly from JayYe’s K cities,then choose v randomly from JayYe’s K cities,so
the case u equals to v is possible).
Suppose you are the king,please determine the K cities so that JayYe is happy.
Because the author is lazy,you only need tell me the minimum expect distance.
Input
The first line contains a single integer T,indicating the number of test cases.
Each test case begins with two integers n and K,indicating the number of cities in this country,the number of cities the king gives to the knight.Then follows n-1 lines,each line contains three integers a,b,and c, indicating there is a road connects city a
and city b with length c.
[Technical Specification]
1 <= T <= 100
1 <= K <= min(50,n)
1 <= n <= 2000
1 <= a,b <= n
0 <= c <= 100000
Output
For each case, output one line, contain one integer, the minimum expect distance multiply K2.
Sample Input
1 2 2 1 2 1
Sample Output
2
/**
hdu 5148 树形dp+分组背包
题目大意：选出K个点v1,v2,...vK使得∑Ki=1∑Kj=1dis(vi,vj)最小.
解题思路：
考虑每条边的贡献，一条边会把树分成两部分，若在其中一部分里选择了x个点，则这条边被统计的次数为x*(K-x)*2.
那么考虑dp[u][i]表示在u的子树中选择了i个点的最小代价，有转移dp[u][i]=minKj=0(dp[u][i?j]+dp[v][j]+j*(K?j)*2*wu,v)，式子中u为v的父亲，wu,v表示(u,v)这条边的长度.
把选取的点看做背包容量，每个子树看做一类物品，从每一类中最多选取一件物品，问背包可获得的最少价值。
时间复杂度O(nK^2).
*/
#include <stdio.h>
#include <string.h>
#include <algorithm>
#include <iostream>
using namespace std;
const int N=2055;
const long long inf = 200000 * 10000000LL;
int n,k;
struct note
{
int v,w,next;
}edge[N*2];
int head[N],ip;
void init()
{
memset(head,-1,sizeof(head));
ip=0;
}
void addedge(int u,int v,int w)
{
edge[ip].v=v,edge[ip].w=w,edge[ip].next=head[u],head[u]=ip++;
}
long long dp[N][55];
void dfs(int u,int pre)
{
for(int i=head[u];i!=-1;i=edge[i].next)
{
int v=edge[i].v;
if(v==pre) continue;
dfs(v,u);
for(int j=k;j>=0;j--)
{
for(int l=1;l<=j;l++)
{
dp[u][j]=min(dp[u][j],dp[u][j-l]+dp[v][l]+(k-l)*l*2*edge[i].w);
}
}
}
}
int main()
{
int T;
scanf("%d",&T);
while(T--)
{
init();
scanf("%d%d",&n,&k);
for(int i=0;i<n-1;i++)
{
int u,v,w;
scanf("%d%d%d",&u,&v,&w);
addedge(u,v,w);
addedge(v,u,w);
}
///由于去最小，初始化除了dp[i][0]和dp[i][1]都要设为无穷大
memset(dp,0,sizeof(dp));
for (int i = 1; i <= n; i++)
for (int j = 2; j <= k; j++)
dp[i][j] = inf;
dfs(1,-1);
printf("%I64d\n",dp[1][k]);
}
return 0;
}