[CF1494D] Dogeforces

[CF1494D] Dogeforces - 分治,LCA

Description

有一棵树，满足每个非叶子结点至少有两个儿子。每个点有权值，满足父亲的权值严格大于儿子的权值。已知这棵树有 (n) 个叶子，并给出这 (n) 个叶子两两之间 ( ext{lca}) 的权值。求出这棵树的结点数，根，形态，以及每个点的权值。

Solution

一种思路是从下往上，排序后用并查集，类似哈夫曼树的构造

一种思路是从上往下，维护当前叶子集合，每次找到一个根，然后将叶子拆成若干份（各个子树），递归处理

怎么找根？任选一个叶子，和他的 LCA 权值最大的那个就是值就是根的权值

怎么划分？LCA 不是根的就是本子树。每次挑一个没划分过的，把 LCA 不是根的找出来

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

#define int long long

const int N = 1005;

int n;
int a[N][N];

// Ans
int ind;
int val[N], fa[N];

// Divide the leaves into subtrees,
// return the id of root
int divide(vector<int> leaves)
{
    // Boundary
    if (leaves.size() == 1)
    {
        val[leaves[0]] = a[leaves[0]][leaves[0]];
        return leaves[0];
    }

    // Find root id
    int max_val = 0, max_id = -1;
    for (auto i : leaves)
    {
        if (a[i][leaves[0]] > max_val)
        {
            max_val = a[i][leaves[0]];
            max_id = i;
        }
    }
    int root = ++ind;
    val[root] = max_val;

    // Divide into subtrees and solve
    vector<int> vis(leaves.size());
    for (int i = 0; i < leaves.size(); i++)
    {
        if (vis[i])
            continue;
        // Build a new subtree
        vector<int> subtree_leaves;
        for (int j = i; j < leaves.size(); j++)
        {
            if (a[leaves[i]][leaves[j]] < max_val)
            {
                // Find new leaf in this subtree
                subtree_leaves.push_back(leaves[j]);
                vis[j] = 1;
            }
        }
        int subtree_root = divide(subtree_leaves);
        fa[subtree_root] = root;
    }
    return root;
}

signed main()
{
    ios::sync_with_stdio(false);
    cin >> n;
    for (int i = 1; i <= n; i++)
    {
        for (int j = 1; j <= n; j++)
        {
            cin >> a[i][j];
        }
    }
    ind = n;
    vector<int> vec;
    for (int i = 1; i <= n; i++)
        vec.push_back(i);

    int root = divide(vec);

    cout << ind << endl;
    for (int i = 1; i <= ind; i++)
        cout << val[i] << " ";
    cout << endl;

    cout << root << endl;
    for (int i = 1; i <= ind; i++)
        if (fa[i])
            cout << i << " " << fa[i] << endl;
}