2017多校训练Contest3: 1005 RXD and dividing hdu6060

最新推荐文章于 2018-11-22 20:08:26 发布

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最新推荐文章于 2018-11-22 20:08:26 发布

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本文链接：https://blog.youkuaiyun.com/lqybzx/article/details/77249085

本文介绍了一种解决特定最小斯坦纳树问题的方法。该问题要求将一系列节点划分成若干组，并计算连接每组中所有节点的最小代价。通过为每组节点分配一个标识并利用DFS遍历树形结构，可以有效地计算出整体最小代价。

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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

Sample Output

bestcoder官方博客:

把1看成整棵树的根. 问题相当于把 $2\sim n$ 每个点一个 $[1, k]$ 的标号. 然后根据最小斯坦纳树的定义, $fa_x)$ 这条边的贡献是 x 子树内不同标号的个数目 $dif_i$ . 那么显然有 $dif_i\leq min(k, sz_i)$ , $sz_i$ 表示子树大小. 可以通过构造让所有 $dif_i$ 都取到最大值. 所以答案就是 $\sum_{x = 2}^{n}{w[x][fa_x] * min(sz_x, k)}$ 时间复杂度 $O (n)$ .

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<cassert>
#include<iostream>
#include<algorithm>
using namespace std;
struct line
{
    int s,t;
    long long x;
    int next;
}a[2000001];
int head[1000001];
int edge;
inline void add(int s,int t,long long x)
{
    a[edge].next=head[s];
    head[s]=edge;
    a[edge].s=s;
    a[edge].t=t;
    a[edge].x=x;
}
bool v[1000001];
long long ans;
int m;
inline int dfs(int d)
{
    v[d]=true;
    int i;
    int sum=1;
    for(i=head[d];i!=0;i=a[i].next)
    {
        int t=a[i].t;
        if(!v[t])
        {
            int xx=dfs(t);
            sum+=xx;
            xx=min(xx,m);
            ans+=(long long)xx*a[i].x;
        }
    }
    return sum;
}
int main()
{
    int n;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        edge=0;
        memset(head,0,sizeof(head));
        int i,s,t,x;
        for(i=1;i<=n-1;i++)
        {
            scanf("%d%d%d",&s,&t,&x);
            edge++;
            add(s,t,x);
            edge++;
            add(t,s,x);
        }
        memset(v,false,sizeof(v));
        ans=0;
        dfs(1);
        printf("%lld\n",ans);
    }
    return 0;
}