E - RXD and dividing
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 SSconnected, 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 4 1 2 3 2 3 4 2 4 5 2 5 6
Sample Output
27
思路:有点构造的感觉 ，就是取一种万能的构造方法！从点1，到所有点遍历一遍就可以了，维护一下 子树节点的个数就可以，再就是k和节点个数关系不同是，转化一下思维就可以，感觉比较有意思的一个题，当初没有想出来，也是怪可惜的。。。
大佬写的很详细：传送门，
我的代码：
#include<bits/stdc++.h>
typedef long long ll;
using namespace std;
struct AA
{
ll v,w,next;
}pos[20000010];
ll f[10000010],num,z;
ll ans,dis[10000010],k;
ll n,u,v,w;
void dfs(ll rt,ll x,ll faa)
{
dis[rt]=1;//cout<<rt<<" "<<x<<endl;
for(int i=f[rt];i!=-1;i=pos[i].next)
{
ll vv=pos[i].v;
if(vv==faa) continue;
dfs(vv,pos[i].w,rt);
dis[rt]+=dis[vv];
}
ans+=x*min(k,dis[rt]);
// cout<<ans<<" "<<x<<" "<<dis[rt]<<" "<<k<<endl;
return;
}
int main()
{
while(scanf("%lld%lld",&n,&k)!=EOF)
{
num=0;
for(int i=0;i<n+4;i++)
{
f[i]=-1;dis[i]=0;
}
for(int i=1;i<n;i++)
{
scanf("%lld%lld%lld",&u,&v,&w);
pos[++num].v=v;
pos[num].next=f[u];
pos[num].w=w;
f[u]=num;
pos[++num].v=u;
pos[num].next=f[v];
pos[num].w=w;
f[v]=num;
}
ans=0;
dfs(1,0,-11);
printf("%lld\n",ans);
}
}
/*
5 4
1 2 3
2 3 4
2 4 5
2 5 6
5 2
1 2 3
2 3 4
2 4 5
2 5 6
*/