[洛谷P4149][IOI2011]Race

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;
}