题目大意：给一棵树，每条边有边权。求一条简单路径，权值和等于$K$，且边的数量最小。
题解：点分治，考虑到这是最小值，不满足可减性，于是点分中的更新答案的地方计算重复的部分要做更改，就用一个数组记录前面的答案。更新答案的时候只从已经访问过的部分来转移。
卡点：一个地方没有$return$，导致$RE$
C++ Code：
#include <cstdio>
#define maxn 200010
#define maxk 1000010
const int inf = 0x3f3f3f3f;
inline int max(int a, int b) {return a > b ? a : b;}
inline int min(int a, int b) {return a < b ? a : b;} int head[maxn], cnt;
struct Edge {
int to, nxt, w;
} e[maxn << 1];
inline void add(int a, int b, int c) {
e[++cnt] = (Edge) {b, head[a], c}; head[a] = cnt;
e[++cnt] = (Edge) {a, head[b], c}; head[b] = cnt;
} bool vis[maxn];
namespace Center_of_Gravity {
int sz[maxn], __nodenum;
int root, MIN;
#define n __nodenum
void __getroot(int u, int fa) {
sz[u] = 1;
int MAX = 0;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa && !vis[v]) {
__getroot(v, u);
sz[u] += sz[v];
MAX = max(MAX, sz[v]);
}
}
MAX = max(MAX, n - sz[u]);
if (MAX < MIN) MIN = MAX, root = u;
}
int getroot(int u, int nodenum = 0) {
n = nodenum ? nodenum : sz[u];
MIN = inf;
__getroot(u, 0);
return root;
}
#undef n
}
using Center_of_Gravity::getroot; int n, k, ans = inf;
int mindep[maxk], W[maxn], dep[maxn], tot;
void getlist(int u, int fa, int val, int __dep) {
if (val <= k) ans = min(ans, __dep + mindep[k - val]);
else return ;
W[++tot] = val, dep[tot] = __dep;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (v != fa && !vis[v]) getlist(v, u, val + e[i].w, __dep + 1);
}
}
inline void init() {
W[tot = 1] = 0;
dep[tot] = 1;
mindep[0] = 0;
}
void solve(int u) {
vis[u] = true;
init();
for (int i = head[u], now = 2; i; i = e[i].nxt) {
int v = e[i].to;
if (!vis[v]) {
getlist(v, u, e[i].w, 1);
while (now <= tot) {
mindep[W[now]] = min(mindep[W[now]], dep[now]);
now++;
}
}
}
for (int i = 1; i <= tot; i++) mindep[W[i]] = inf;
for (int i = head[u]; i; i = e[i].nxt) {
int v = e[i].to;
if (!vis[v]) {
solve(getroot(v));
}
}
} int main() {
scanf("%d%d", &n, &k);
for (int i = 1, a, b, c; i < n; i++) {
scanf("%d%d%d", &a, &b, &c); a++, b++;
add(a, b, c);
}
__builtin_memset(mindep, 0x3f, sizeof mindep);
solve(getroot(1, n));
if (ans != inf) printf("%d\n", ans);
else puts("-1");
return 0;
}