#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 100010, M = N * 2;
int n;
int h[N], e[M], ne[M], idx;
bool st[N];
int ans = N;
void add(int a, int b)
{
e[idx] = b, ne[idx] = h[a], h[a] = idx ++;
}
// 以u为根的子树中点的数量
int dfs(int u)
{
st[u] = true; // 标记一下,已经被搜过了
int sum = 1, res = 0; //sum 当前子树的大小, res 把这个点删除后每一个连通块大小的最大值
for(int i = h[u]; i != -1; i = ne[i])
{
int j = e[i];
if(!st[j])
{
int s = dfs(j); // 当前子树的大小
res = max(res, s);
sum += s;
}
}
res = max(res, n - sum);
ans = min(ans, res);
return sum;
}
int main()
{
cin >> n;
memset(h, -1, sizeof h);
for(int i = 0; i < n - 1; i++)
{
int a, b;
cin >> a >> b;
add(a, b), add(b, a);
}
dfs(1);
cout << ans << endl;
return 0;
}
标签:优先,遍历,idx,int,res,sum,ans,ne,深度
From: https://blog.51cto.com/u_16492348/9634930