PAT 真题 | 1021 Deepest Root

近期在刷 PAT 题库的时候遇到一个求树的最长根的问题, 觉得证明挺有意思的, 于是抽空记录下来.

1021 Deepest Root（25 分）

A graph which is connected and acyclic can be considered a tree. The hight of the tree depends on the selected root. Now you are supposed to find the root that results in a highest tree. Such a root is called the deepest root.

Input Specification:

Each input file contains one test case. For each case, the first line contains a positive integer N (≤10^4) which is the number of nodes, and hence the nodes are numbered from 1 to N. Then N−1 lines follow, each describes an edge by given the two adjacent nodes’ numbers.

Output Specification:

For each test case, print each of the deepest roots in a line. If such a root is not unique, print them in increasing order of their numbers. In case that the given graph is not a tree, print Error: K components where K is the number of connected components in the graph.

Sample Input 1:

5
1 2
1 3
1 4
2 5

Sample Output 1:

3
4
5

Sample Input 2:

5
1 3
1 4
2 5
3 4

Sample Output 2:

Error: 2 components

题目分析

这道题首先要求我们判断输入的图能否构成一棵树. 如果能构成一棵树就按节点序号顺序输出这棵树的最长根节点, 最长根节点是能让一棵树的深度达到最大值的节点. 如果不能构成一棵树就输出错误, 并报告图的联通部分个数.

在确定能构成了一棵树后, 使用两次 dfs 或 bfs 搜索算法就能找到所有的最长根节点. 下面证明这个结论:

假设树的最大深度为 maxD. 首先以任意一点作为树的根节点, 假设第一次搜索出的叶子节点为 L = {L1, L2, …, Ln}. 对应的节点深度为 {A1, A2, …, An}. 节点深度的最大值为 maxA, 对应的节点集合假设为 {Lmax}. 从 L 中任选两个叶子节点的组合, 所有组合之间的距离最大值就是该树所能达到的最大深度 maxD.
若找到了 i 和 j , 使得 maxD = Li 和 Lj 对应的深度, 不妨设 Ai >= Aj.
若Li和Lj不在同一支上, 则Li和Lj对应的深度 = Ai + Aj
另一方面, Ai + Aj <= maxA + Aj <= maxD.
所以, Ai + Aj = maxA + Aj = maxD, maxA = Ai, Ai 属于 {Lmax}. 根据定义, {Lmax} 集合中的元素为最长根节点.

第二次搜索从 {Lmax} 中任选一个元素, 即选择一个已经找到的最长根节点 D1. 根据定义, 以该点为根形成的树的最大深度是 maxD. 对应最大深度的节点假设为 Di, 那么反过来以 Di 为根产生的树的最大深度也是 maxD. 所以 Di 也是最长根节点.接下来还要证明已经找出了所有的最长根节点.

实例代码

#include<stdio.h>
#include<string.h>
#include<vector>
#include<set>
using namespace std;

#define For(i,n) for(int i=0; i<n; ++i)
const int maxn = 10010;
vector<int> G[maxn];
int root[maxn]; // 并查集
set<int> deepestroot;
set<int> temproot;
bool done[maxn]; // dfs标记

void initSet(int n){
    for(int i = 1; i <= n; i++) root[i] = i;
}
int findSet(int x){
    return root[x] == x ? x : root[x]=findSet(root[x]);
}

void dfs(int root, int height){
    static int maxheight = 0;
    if (done[root]) return ;
    done[root] = true;
    For(i, G[root].size()){
        dfs(G[root][i], height+1);
    }
    if (height > maxheight){
        maxheight = height;
        temproot.clear();
        temproot.insert(root);
    } else if (height == maxheight){
        temproot.insert(root);
    }
}

int main()
{
    int N;
    scanf("%d", &N);
    initSet(N);
    For(i,N-1){
        int a,b;
        scanf("%d%d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
        int x = findSet(a), y = findSet(b);
        if (x != y) root[x] = y;
    }

    set<int> components;
    for(int i = 1; i <= N; ++i)
        components.insert(findSet(i));
    if (components.size() > 1){
        printf("Error: %ld components\n", components.size());
    } else {
        memset(done, false, sizeof done);
        dfs(1, 0);
        deepestroot = temproot;
        memset(done, false, sizeof done);
        dfs(*temproot.begin(), 0);
        for (auto it=temproot.begin(); it!=temproot.end(); ++it) deepestroot.insert(*it);
        for (auto it=deepestroot.begin(); it!=deepestroot.end(); ++it) printf("%d\n", *it);
    }
    return 0;
}