题面
题解
根据题意，不回头是最好的（显然法）
\(dfs\)找到最长链，设长度为\(\mathrm{L}\)，然后分类讨论：
如果\(\mathrm{L} > m\)，答案就是\(m + 1\)
否则显然可以多走\(m-\mathrm{L} + 1\)步，可以多访问\((m-\mathrm{L} + 1) / 2\)步，于是答案就是
\(min\{n, \mathrm{L} + (m-\mathrm{L} + 1) / 2\}\)
证明？？？这一辈子都不可能的
代码
#include
#include
#include
#include
#define RG register
#define file(x) freopen(#x".in", "r", stdin), freopen(#x".out", "w", stdout)
#define clear(x, y) memset(x, y, sizeof(x))
inline int read()
{
int data = 0, w = 1; char ch = getchar();
while(ch != '-' && (!isdigit(ch))) ch = getchar();
if(ch == '-') w = -1, ch = getchar();
while(isdigit(ch)) data = data * 10 + (ch ^ 48), ch = getchar();
return data * w;
}
const int N(110);
struct edge { int next, to; } e[N << 1];
int head[N], e_num, n, m, maxdep, vis[N];
inline void add_edge(int from, int to)
{
e[++e_num] = (edge) {head[from], to};
head[from] = e_num;
}
void dfs(int x, int dep)
{
maxdep = std::max(maxdep, dep), vis[x] = 1;
for(RG int i = head[x]; i; i = e[i].next)
{
int to = e[i].to; if(vis[to]) continue;
dfs(to, dep + 1);
}
}
int main()
{
#ifndef ONLINE_JUDGE
file(cpp);
#endif
n = read(), m = read();
for(RG int i = 1, a, b; i < n; i++)
a = read(), b = read(), add_edge(a, b), add_edge(b, a);
dfs(0, 1);
if(m < maxdep) printf("%d\n", m + 1);
else printf("%d\n", std::min(n, maxdep + (m - maxdep + 1) / 2));
return 0;
}