HDU 6060 RXD and dividing （最小斯坦纳树）

RXD and dividing

题目链接

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

题意：
有n个点，其中1为起点，其余的n-1个点（2~n)为我们要到达的点，将剩余的这n-1个点分成k个集合，看一下到这些点的路径的长度。
分析： 根据最小斯坦纳树的定义，（就算没有这个定义，我们也应该能够想到这一点，如果要想让路径尽可能的长，那么对于一个节点即它的所有子树来说，最多能够被分成k部分）一条边（u,father(u))对整个的贡献就相当于这条边的value值乘上k,与size [ x ] (当前情况下x的子节点个数，包括它本身)中的较小者，我们只要遍历所有的边，然后将每条边的值都算出来，最后求和。

#include<bits/stdc++.h>
using namespace std;

const int N=1e6+7;
int head[N],nxt[N*2],v[N*2],w[N*2],sz[N];
int n,k,g[N],ed;
long long ans;
inline void adg(int x,int y,int z)///头插法
{
    v[++ed]=y;///v表示的是终点
    w[ed]=z;///w表示的是这条边的权重
    nxt[ed]=head[x];///xt表示的是钱一条边
    head[x]=ed;///头的指向改变
}

void dfs(int x,int fa,int val)
{
    sz[x]=1;
    for(int i=head[x]; i!=0; i=nxt[i])
        if(v[i]!=fa)///不是又找到本身这条边了
        {
            dfs(v[i],x,w[i]);
            sz[x]+=sz[v[i]];
        }
    ans+=1ll*min(k,sz[x])*val;
}

int main()
{
    while(~scanf("%d%d",&n,&k))
    {
        for(int i=1; i<=n; i++)
            head[i]=0;///每一个点的指向都赋初值
        ed=0;
        ans=0;
        for(int i=1;i<n;i++)
        {
            int x,y,z;
            scanf("%d%d%d",&x,&y,&z);
            adg(x,y,z),adg(y,x,z);
        }
        dfs(1,0,0);
        printf("%lld
",ans);
    }
    return 0;
}