题意
给你一颗有根树，你要选择(k)个点，最大化(sum_{i in S} val_i)，其中(S)是被选点的集合，(val_i)等于节点(i)到根的路径上未被选择点的个数。
解题思路
这题思路挺有意思的。
我们可以逐个点进行选择。新选择一个点对答案的贡献为点到跟的距离减去子树中被选择点的个数。
这样子的话，很容易可以得出，如果点(u)在答案中，那么它所有的后代也必须在答案中。
由此，我们可以为每个点赋一个权值(v_i = d_i - sub_i)，(d_i)表示点(i)到根的距离，(sub_i)表示点(i)后代的个数。
AC代码
#include <bits/stdc++.h>
using namespace std;
// #include <ext/rope>
// using namespace __gnu_cxx;
// #include <ext/pb_ds/assoc_container.hpp>
// #include <ext/pb_ds/tree_policy.hpp>
// using namespace __gnu_pbds;
// typedef ll key_type;
// typedef null_mapped_type value_type;
// typedef tree<key_type, value_type, less<key_type>, rb_tree_tag, tree_order_statistics_node_update> rbtree;
// typedef __gnu_pbds::priority_queue<pi,greater<pi>,pairing_heap_tag > heap;
// mt19937 rng(chrono::steady_clock::now().time_since_epoch().count());
// int rnd(int l,int r){return l+rng()%(r-l+1);}
typedef long long ll;
typedef double db;
typedef pair<int,int> PI;
typedef vector<int> VI;
#define rep(i,_,__) for (int i=_; i<=__; ++i)
#define per(i,_,__) for (int i=_; i>= __; --i)
#define pb push_back
#define mp make_pair
#define fi first
#define se second
#define x1 _x
#define x2 __x
#define y1 _y
#define y2 __y
#define SZ(x) ((int)(x).size())
#define all(x) (x).begin(),(x).end()
#define rall(x) (x).rbegin(),(x).rend()
#define endl '
'
const double pi = acos(-1.0);
namespace IO{
bool REOF = 1; //为0表示文件结尾
inline char nc() {
static char buf[1 << 20], *p1 = buf, *p2 = buf;
return p1 == p2 && REOF && (p2 = (p1 = buf) + fread(buf, 1, 1 << 20, stdin), p1 == p2) ? (REOF = 0, EOF) : *p1++;
}
template<class T>
inline bool read(T &x) {
char c = nc();bool f = 0; x = 0;
while (c<'0' || c>'9')c == '-' && (f = 1), c = nc();
while (c >= '0'&&c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = nc();
if(f)x=-x;
return REOF;
}
template<class T>
inline bool write(T x){
if(x > 9) write(x / 10);
putchar('0'+x%10);
return REOF;
}
template<typename T, typename... T2>
inline bool read(T &x, T2 &... rest) {
read(x);
return read(rest...);
}
inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')); }
// inline bool need(char &c) { return ((c >= 'a') && (c <= 'z')) || ((c >= '0') && (c <= '9')) || ((c >= 'A') && (c <= 'Z')) || c==' '; }
inline bool read_str(char *a) {
while ((*a = nc()) && need(*a) && REOF)++a; *a = ' ';
return REOF;
}
inline bool read_db(double &x){
bool f = 0; char ch = nc(); x = 0;
while(ch<'0'||ch>'9') {f|=(ch=='-');ch=nc();}
while(ch>='0'&&ch<='9'){x=x*10.0+(ch^48);ch=nc();}
if(ch == '.') {
double tmp = 1; ch = nc();
while(ch>='0'&&ch<='9'){tmp=tmp/10.0;x=x+tmp*(ch^48);ch=nc();}
}
if(f)x=-x;
return REOF;
}
template<class TH>
inline void _dbg(const char *sdbg, TH h){ cerr<<sdbg<<'='<<h<<endl; }
template<class TH, class... TA>
inline void _dbg(const char *sdbg, TH h, TA... a) {
while(*sdbg!=',')cerr<<*sdbg++;
cerr<<'='<<h<<','<<' '; _dbg(sdbg+1, a...);
}
template<class T>
ostream &operator<<(ostream& os, vector<T> V) {
os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
}
template<class T>
ostream &operator<<(ostream& os, set<T> V) {
os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
}
template<class T>
ostream &operator<<(ostream& os, map<T,T> V) {
os << "[ "; for (auto vv : V) os << vv << ","; return os << " ]";
}
template<class L, class R>
ostream &operator<<(ostream &os, pair<L,R> P) {
return os << "(" << P.x << "," << P.y << ")";
}
#ifdef BACKLIGHT
#define debug(...) _dbg(#__VA_ARGS__, __VA_ARGS__)
#else
#define debug(...)
#endif
}
using namespace IO;
const int N = 2e5 + 5;
const int M = 2e5 + 5;
const int MAXV = 1e6 + 5;
const int MOD = 1e9+7; // 998244353 1e9+7
const int INF = 0x3f3f3f3f; // 1e9+7 0x3f3f3f3f
const ll LLINF = 0x3f3f3f3f3f3f3f3f; // 1e18+9 0x3f3f3f3f3f3f3f3f
const double eps = 1e-8;
// int dx[4] = { 0, 1, 0, -1 };
// int dx[8] = { 1, 0, -1, 1, -1, 1, 0, -1 };
// int dy[4] = { 1, 0, -1, 0 };
// int dy[8] = { 1, 1, 1, 0, 0, -1, -1, -1 };
// ll qp(ll a, ll b) {
// ll res = 1;
// a %= mod;
// assert(b >= 0);
// while(b){
// if(b&1)
// res = res * a % mod;
// a = a * a % mod;
// b >>= 1;
// }
// return res;
// }
// ll inv(ll x) {return qp(x, mod - 2);}
// ll factor[N], finv[N];
// void init() {
// factor[0]=1;
// for(int i=1; i<N; i++) factor[i] = factor[i-1] * i % mod;
// finv[N-1] = qp(factor[N-1], mod - 2);
// for(int i=N-2; i>=0; i--) finv[i] = finv[i+1] * (i+1) % mod;
// }
// ll c(ll n, ll m) {
// return factor[n] * finv[m] % mod * finv[n-m] % mod;
// }
// #define ls (x<<1)
// #define rs (x<<1|1)
// #define mid ((l+r)>>1)
// #define lson ls,l,mid
// #define rson rs,mid+1,r
#define fore(_, __) for(int _ = head[__]; _; _=e[_].nxt)
int head[N], tot = 1;
struct Edge {
int v, nxt;
Edge(){}
Edge(int _v, int _nxt):v(_v), nxt(_nxt) {}
}e[N << 1];
void addedge(int u, int v) {
e[tot] = Edge(v, head[u]); head[u] = tot++;
e[tot] = Edge(u, head[v]); head[v] = tot++;
}
/**
* ********** Backlight **********
* 仔细读题
* 注意边界条件
* 记得注释输入流重定向
* 没有思路就试试逆向思维
* 我不打了，能不能把我的分还给我
*/
int d[N], sz[N];
void dfs(int u, int f) {
d[u] = d[f] + 1;
sz[u] = 1;
fore(i, u) {
int v = e[i].v;
if(v==f)continue;
dfs(v, u);
sz[u] += sz[v];
}
}
int n, k, a[N];
void solve(int Case) {
read(n, k);
int u, v;
rep(i, 1, n-1) {
read(u, v);
addedge(u, v);
}
dfs(1, 0);
rep(i, 1, n) a[i] = d[i] - sz[i];
sort(a+1, a+1+n);
ll ans = 0;
rep(i, 1, k) ans += (ll)a[n-i+1];
printf("%lld
", ans);
}
int main()
{
#ifdef BACKLIGHT
freopen("in.txt", "r", stdin);
#endif
// ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int _T; read(_T); for (int _ = 1; _ <= _T; _++) solve(_);
// int _T=1; while(read(n)) solve(_T), _T++;
solve(1);
return 0;
}