2017杭电多校联赛第三场-RXD and dividing （hdu6060）最小生成树

RXD and dividing

Time Limit: 6000/3000 MS (Java/Others)    Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 807    Accepted Submission(s): 339

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

题目大意：给出一颗n个节点的树，要求将2-n号节点分成k部分，然后再将每一部分加上1号节点，定义每一部分的val为其中的点在原图上的最小斯坦纳树，问总的val最大可能是多少。

解题思路：我们可以先将1看成整颗树的根，然后将余下的2-n号节点分为k部分，具体怎么分呢，我们可以将每颗子树的子树分为k部分，如果子树大于k的话，否则就分为子树的数量的部分，目的就是要得到尽可能大的总值，因为在算每一部分的值的时候都需要经过这颗子树和它的父亲，所以我们将每颗子树尽可能多的分块，然后将每一块的值相加就可以了。答案是∑​x=2​n​​w[x][fa​x​​]∗min(sz​x​​,k) 。

代码使用dfs写的。

具体看代码：

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL MOD=1e9+7;
const int MAXN=1e6+7;
int head[MAXN],tot;
int sz[MAXN],f[MAXN],weight[MAXN];

struct Edge
{
    int from,to,cost,nxt;
}e[MAXN*2];

void addedge(int u,int v,int w)
{
    e[tot].from=u;
    e[tot].to=v;
    e[tot].cost=w;
    e[tot].nxt=head[u];
    head[u]=tot++;
}

void dfs(int u,int fa)
{
    sz[u]=1;
    for(int i=head[u];i!=-1;i=e[i].nxt)
    {
        int to=e[i].to;
        if(to==fa)  continue;
        f[to]=u;
        weight[to]=e[i].cost;
        dfs(to,u);
        sz[u]+=sz[to];
    }
}

int main()
{
    int n,k;
    while(scanf("%d%d",&n,&k)!=EOF)
    {
        memset(sz,0,sizeof(sz));
        memset(head,-1,sizeof(head));
        tot=0;
        int u,v,w;
        for(int i=1;i<=n-1;++i)
        {
            scanf("%d%d%d",&u,&v,&w);
            addedge(u,v,w);
            addedge(v,u,w);
        }
        dfs(1,-1);
        LL ans=0;
        for(int i=2;i<=n;i++)
        {
            ans+=(LL)weight[i]*min(sz[i],k);
        }
        printf("%lld\n",ans);
    }
    return 0;
}

题目链接： 点击打开链接http://acm.hdu.edu.cn/showproblem.php?pid=6060