题目描述

blablabla

样例

blablabla

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

#include<iostream>
#include<algorithm>
#include<cstring>
#include<queue>

using namespace std;

const int N = 100010, M = 2 * N;
int n;
int h[N], e[M], ne[M], idx;
bool st[N];

void add(int a, int b)
{
    e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}

int dfs(int root, int father)
{
    int res = 1;
    for (int i = h[root]; ~i; i = ne[i])
    {
        int j = e[i];
        if (j != father) res += dfs(j, root);
    }
    return res;
}

int bfs(int root)
{
    int res = 0;
    queue<int> q;
    q.push(root);
    st[root] = true;
    while(q.size())
    {
        auto t = q.front(); q.pop();
        res ++;
        for (int i = h[t]; ~i; i = ne[i])
        {
            int j = e[i];
            if (!st[j])//入队一次；
            {
                q.push(j);
                st[j] = true;
            }
        }
    }
    return res;
}
int main()
{
    memset(h, -1, sizeof h);
    cin >> n;
    for (int i = 0; i < n - 1; i ++)
    {
        int a, b;
        cin >> a >> b;
        add(a, b);
        add(b, a);
    }
    int res = 0;
    st[1] = true;
    for (int i = h[1]; i != -1; i = ne[i])
    {
        int j = e[i];
        //res = max(res, helper(j, 1));
        res = max(res, bfs(j));
    }
    cout << res;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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