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
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.
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
启发博客：http://blog.csdn.net/lz161530245/article/details/76794473
以下题意和题解摘自此博客
题意：给一棵树T，有n个结点。
给一个k，表示有k个集合，我们需要把2,3,4,…n号节点放入集合，要保证k个集合的并集等于{2,3,4,5…n}， 并且集合互不相交。(集合可以为空)
然后每次取一个集合Si与{1}求并，得到比如{1,2,3}，那么tempi = f({1,2,3})；f({1}并Si)的意思是把集合内的 所有点连接起来的边的权值和。最后把所有权值和相加的到答案。
最后问你能够得到最大的答案。
给一个k，表示有k个集合，我们需要把2,3,4,…n号节点放入集合，要保证k个集合的并集等于{2,3,4,5…n}， 并且集合互不相交。(集合可以为空)
然后每次取一个集合Si与{1}求并，得到比如{1,2,3}，那么tempi = f({1,2,3})；f({1}并Si)的意思是把集合内的 所有点连接起来的边的权值和。最后把所有权值和相加的到答案。
最后问你能够得到最大的答案。
分析：
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我们要想得到最大的答案，那么就要尽可能的去利用这些边，也就是尽可能重复计算这些边。
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那么我们想，假设先从叶子节点开始，把这些叶子节点放入一个集合，那么这个集合的temp值就会把所有的边都算一遍。那么下次我们取所有叶子节点的父亲，放入一个集合，那么这个集合的temp值会把除了叶子节点到父亲的那条那边的其他所有边都算一遍。因为集合可以为空，以此类推，我们就可以得到最大的答案。但是如果遇到集合不够的情况，就把剩下的所有点加入最后一个集合。
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那么有以上分析，其实就是算每条边会算多少次，比如叶子节点到父亲的那条边会算一次。其实一条边会算多少次跟某个点的所有子孙节点个数有关，就比如样例中，2号点有3个子孙节点， 那么2号点连接父节点的那条边会算3+1次。3号点有0个子孙节点，那么3号点连接父节点的那条边会算0+1次。
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那么其实问题就是转化为求每个点的子孙节点个数，然后算出每条边要重复计算的次数即可。
1 #include<cstdio> 2 #include<iostream> 3 #include<vector> 4 #include<cmath> 5 using namespace std; 6 #define MAXN 1000005 7 8 struct Edge 9 { 10 int v,w; 11 };//Edge[u]里面存着u连接的v和这条边的权值 12 vector<Edge>vec[MAXN];//容器存每个u对应的v和w 13 int size[MAXN];//每个节点的子节点数 14 int weight[MAXN];//weight[v]记录v点和它父节点的边值 15 16 17 void dfs(int u,int pre)//u是当前点,pre是父节点 18 { 19 size[u]=1;//它本身 20 int len=vec[u].size(); 21 int vv;//u的子节点 22 for(int i=0;i<len;i++) 23 { 24 vv=vec[u][i].v; 25 weight[vv]=vec[u][i].w;//记录v点和它父节点的边值 26 dfs(vv,u); 27 size[u]+=size[vv]; 28 } 29 } 30 31 int main() 32 { 33 int n,k; 34 while(~scanf("%d%d",&n,&k)) 35 { 36 for(int i=1;i<=n;i++) 37 { 38 vec[i].clear(); 39 size[i]=0; 40 weight[i]=0; 41 } 42 Edge tmp; 43 for(int i=1;i<=n-1;i++) 44 { 45 int u,v,w; 46 scanf("%d%d%d",&u,&v,&w); 47 tmp.w=w; 48 //tmp.v=u; 49 //vec[v].push_back(tmp); 50 tmp.v=v; 51 vec[u].push_back(tmp); 52 } 53 dfs(1,0);//深搜找出一个子树节点的个数 54 long long sum=0; 55 for(int i=2;i<=n;i++)//只分2-n 56 { 57 sum+=(long long)weight[i]*min(size[i],k); 58 } 59 printf("%lld ",sum); 60 } 61 return 0; 62 }
这种写法比较好理解，但我感觉如果遇到不是按照父子顺序给出的权值的话，这种写法就不太适合，邻接表会好一点