HDU 6060 RXD and dividing

RXD and dividing

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 1253    Accepted Submission(s): 527

Problem Description

RXD has a tree  T , with the size of  n . Each edge has a cost.
Define  f(S)  as the the cost of the minimal Steiner Tree of the set  S  on tree  T . 
he wants to divide  2,3,4,5,6,…n  into  k  parts  S1,S2,S3,…Sk ,
where  ⋃Si={2,3,…,n}  and for all different  i,j  , we can conclude that  Si⋂Sj=∅ . 
Then he calulates  res=∑ki=1f({1}⋃Si) .
He wants to maximize the  res .
1≤k≤n≤106
the cost of each edge∈[1,105]
Si  might be empty.
f(S)  means that you need to choose a couple of edges on the tree to make all the points in  S  connected, and you need to minimize the sum of the cost of these edges.  f(S)  is equal to the minimal cost 

 

Input

There are several test cases, please keep reading until EOF.
For each test case, the first line consists of 2 integer  n,k , which means the number of the tree nodes , and  k  means the number of parts.
The next  n−1  lines consists of 2 integers,  a,b,c , means a tree edge  (a,b)  with cost  c .
It is guaranteed that the edges would form a tree.
There are 4 big test cases and 50 small test cases.
small test case means  n≤100 .

 

Output

For each test case, output an integer, which means the answer.

 

Sample Input

  

   5 4
1 2 3
2 3 4
2 4 5
2 5 6
  

 

Sample Output

  

   27
  

 

Source

2017 Multi-University Training Contest - Team 3 

 

题意：

有n个顶点，给你n-1条边，让你在把（2-n）顶点分成k个部分（也可以分成1到k-1个部分），每个部分和树根顶点1组成一颗树，求这k个树的边权制总和最大。

POINT：

n个顶点，和n-1条边，证明给你的树就是最小生成树，那么无论你怎么分，其实都是最小生成树，所以题目里的最小斯坦纳树是没什么用的。

我们要求的是总和最大，我们就让每个边使用次数最多。易得每个边（x和他的父亲节点形成的边）使用次数max=min(k,size[x])。

那么答案就是sum(每个边的权制*min（k,size[x]))。

#include <stdio.h>
#include <iostream>
#include <string.h>
#include <math.h>
using namespace std;
#define  LL long long
const int N = 1e6+5;
int pre[N],head[N];
struct node
{
    int v,nxt,w;
}len[N<<1];
int sum[N];
int cnt;
void add(int u,int v,int w)
{
    cnt++;
    len[cnt].v=v;
    len[cnt].nxt=head[u];
    len[cnt].w=w;
    head[u]=cnt;
}
void dfs(int u,int p)
{
    sum[u]=1;
    for(int i=head[u];i!=-1;i=len[i].nxt)
    {
        if(len[i].v==p) continue;
        pre[len[i].v]=len[i].w;
        dfs(len[i].v,u);
        sum[u]+=sum[len[i].v];
    }
}
int main()
{
    int n,k;
    while(~scanf("%d %d",&n,&k))
    {
        cnt=0;
        memset(head,-1,sizeof head);
        memset(sum,0,sizeof sum);
        memset(pre,0,sizeof pre);
        for(int i=1;i<n;i++)
        {
            int u,v,w;
            scanf("%d %d %d",&u,&v,&w);
            add(u,v,w);
            add(v,u,w);
        }
        LL ans=0;
        dfs(1,-1);
        for(int i=2;i<=n;i++)
        {
            ans+=(LL)(1LL*pre[i]*1LL*min(k,sum[i]));
        }
        printf("%lld\n",ans);
    }
    
}