2017 Multi-University Training Contest 3（ hdu 6060） RXD and dividing

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:把2~n这n-1个点分成k份，然后定义一个f（s）表示要把集合S里点全部变成联通要加边权的和的最小值。

然后问你怎么分这n-1个点才能使每个集合加上只有点1这个集合的f（s）的和最大，求出这个最大值。集合可以为空集。

可以发现，空集是没有什么必要的，如果合并两个集合，必然会使答案变小或者不变，所以我们尽量让点分布到不同的集合中去。

我们可以考虑每一条边的贡献，对于一个点，它与它父亲连的那条边，只有可能会被这个点，还有这个点的后代经过，所以我们想让这个点与它的后代尽可能的分到不同的集合中去，所以这条边的贡献就是min（k，这个点后代的个数加上它本身这个点）。

#include<vector>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define LL long long

using namespace std;

struct edge
{
int to,cost;
edge(int _to,int _cost)
{
to = _to;
cost = _cost;
}
};

vector<edge> e[1000010];
int n,k;
LL ans;

int dfs(int x,int pre,int PreEdge)
{
int sz = 1;
for(int i=0;i<e[x].size();i++)
{
int xx = e[x][i].to;
if(xx == pre)
continue;
sz += dfs(xx,x,e[x][i].cost);
}
ans += (LL)PreEdge*min(k,sz);
return sz;
}

int main(void)
{
int T,i,j;
while(scanf("%d%d",&n,&k)==2)
{
for(i=1;i<=n;i++)
e[i].clear();
for(i=1;i<=n-1;i++)
{
int x,y,z;
scanf("%d%d%d",&x,&y,&z);
e[x].push_back(edge(y,z));
e[y].push_back(edge(x,z));
}
ans = 0;
dfs(1,-1,0);
printf("%lld\n",ans);
}

return 0;
}