hdu 6060 RXD and dividing (贪心）

RXD has a tree TT, with the size of nn. Each edge has a cost. 
Define f(S)f(S) as the the cost of the minimal Steiner Tree of the set SS on tree TT. 
he wants to divide 2,3,4,5,6,…n2,3,4,5,6,…n into kk parts S1,S2,S3,…SkS1,S2,S3,…Sk, 
where ⋃Si={2,3,…,n}⋃Si={2,3,…,n} and for all different i,ji,j , we can conclude that Si⋂Sj=∅Si⋂Sj=∅. 
Then he calulates res=∑ki=1f({1}⋃Si)res=∑i=1kf({1}⋃Si). 
He wants to maximize the resres. 
1≤k≤n≤1061≤k≤n≤106 
the cost of each edge∈[1,105]the cost of each edge∈[1,105] 
SiSi might be empty. 
f(S)f(S) means that you need to choose a couple of edges on the tree to make all the points in SS connected, and you need to minimize the sum of the cost of these edges. f(S)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,kn,k, which means the number of the tree nodes , and kk means the number of parts. 
The next n−1n−1 lines consists of 2 integers, a,b,ca,b,c, means a tree edge (a,b)(a,b) with cost cc. 
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≤100n≤100.

Output

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

Sample Input

5 41 2 32 3 42 4 52 5 6

Sample Output

27

给你一个树n个数，分成m组，每组的最小生成树权值加起来的总和最小

其实是个水题。。队友想复杂了。。（都想到拓扑了）。最后自己出坑发现正确思路还是很开心的

（ε=ε=ε=┏(゜ロ゜;)┛逃  算每个点能用的次数 和k比较 取小的，就是能用的点尽可能贡献

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int N = 1e6+100;int sze[N];typedef pair<int,ll> P;vector<P> e[N];int n,m;ll ans=0;void dfs(int x,int y,ll w){sze[x]=1;for(int i=0;i<e[x].size();i++){if(e[x][i].first==y) continue;dfs(e[x][i].first,x,e[x][i].second);sze[x]+=sze[e[x][i].first];}ans+=1LL*min(sze[x],m)*w;}int main(){while(scanf("%d%d",&n,&m)!=EOF){ans=0;for(int i=1;i<=n;i++)e[i].clear();for(int i=1;i<n;i++){int a,b;ll c;scanf("%d%d%lld",&a,&b,&c);e[a].push_back(P(b,c));e[b].push_back(P(a,c));}dfs(1,0,0);printf("%lld\n",ans );}}