Balancing Act （树形dp+树的重心）

最新推荐文章于 2021-07-22 10:22:48 发布

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最新推荐文章于 2021-07-22 10:22:48 发布

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分类专栏：动态规划树形dp 文章标签：动态规划树形dp 算法

本文链接：https://blog.youkuaiyun.com/a1046765624/article/details/78555536

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本文介绍了一种通过深度优先搜索(DFS)算法来寻找一棵树的重心的方法，并提供了一个具体的实现案例。文章详细解释了树的重心概念及其计算过程，旨在帮助读者理解如何找到树中具有最小平衡度的节点。

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Consider a tree T with N (1 <= N <= 20,000) nodes numbered 1...N. Deleting any node from the tree yields a forest: a collection of one or more trees. Define the balance of a node to be the size of the largest tree in the forest T created by deleting that node from T.
For example, consider the tree:

Deleting node 4 yields two trees whose member nodes are {5} and {1,2,3,6,7}. The larger of these two trees has five nodes, thus the balance of node 4 is five. Deleting node 1 yields a forest of three trees of equal size: {2,6}, {3,7}, and {4,5}. Each of these trees has two nodes, so the balance of node 1 is two.

For each input tree, calculate the node that has the minimum balance. If multiple nodes have equal balance, output the one with the lowest number.
Input

The first line of input contains a single integer t (1 <= t <= 20), the number of test cases. The first line of each test case contains an integer N (1 <= N <= 20,000), the number of congruence. The next N-1 lines each contains two space-separated node numbers that are the endpoints of an edge in the tree. No edge will be listed twice, and all edges will be listed.

Output

For each test case, print a line containing two integers, the number of the node with minimum balance and the balance of that node.

Sample Input

Sample Output

1 2

题目大概：

给出一棵树，找出它的重心是哪个节点，它的最小额度是多少（即最大子树的结点数）。

思路：

首先要知道重心定义，

树的重心是，与重心相连的所有子树中，结点数最多的子树是所有节点的最多子树中最少的。

然后就dfs遍求出最小的额度。然后从1开始判断就可以了。

代码：

#include <iostream>
#include <cstdio>
#include <cstring>
using namespace std;

int n;
int dp[20005];
int d[20005];
int head[20005];
int minshu,no;
int ans;
struct shu
{
    int v;
    int next;
}tr[40010];

void add(int q,int w)
{
    tr[ans].v=w;
    tr[ans].next=head[q];
    head[q]=ans++;

}

void dfs(int x,int pa)
{
    d[x]=1;
    int maxshu=-20005;
    for(int i=head[x];i!=-1;i=tr[i].next)
    {
        int son=tr[i].v;
        if(son!=pa)
        {
            dfs(son,x);
            d[x]+=d[son];
            maxshu=max(maxshu,d[son]);
        }
    }
    dp[x]=max(maxshu,n-d[x]);
    if(minshu>dp[x])
    {
        minshu=dp[x];

    }

}

int main()
{
    int t;
    scanf("%d",&t);
   while(t--)
   {
       memset(dp,0,sizeof(dp));
       memset(d,0,sizeof(d));
       memset(head,-1,sizeof(head));
       ans=0;
       scanf("%d",&n);
       for(int i=1;i<n;i++)
       {
           int q,w;
            scanf("%d%d",&q,&w);
            add(q,w);
            add(w,q);

       }
       minshu=20005;
       no=0;
       dfs(1,0);
       for(int i=n;i>=1;i--)
       {
           if(minshu==dp[i])
           {
               no=i;
           }
       }
       printf("%d %d\n",no,minshu);
   }
    return 0;
}