洛谷P1099 树网的核

(Large extbf{Description: } large{给你一个无根树和一个非负整数 ext{s}，求直径上的一段长度不超过 ext{s}的路径F，使树上结点到F距离最大值的最小值。(5 leq n leq 300, 0 leq s leq 1000)\})

(Large extbf{Solution: } large{一道特别经典的题目，解法很多，值得反复推敲。\介绍一种解法。我们可以先 ext{dfs}找出树的直径，并在 ext{dfs}过程中维护 ext{dis}与 ext{fa}。\我们可以在直径上借助尺取法，遍历一遍直径顺便维护 ext{ans}。\注意，当 ext{s}大于直径的时候，我们的最大值就要 ext{dfs}一下，求出其余点到直径上任意一点的距离的最大值即可。})

(Large extbf{Code: })

#include <cstdio>
#include <algorithm>
#define LL long long
#define gc() getchar()
#define rep(i, a, b) for(int i = (a); i <= (b); ++i)
#define _rep(i, a, b) for(int i = (a); i <= (b); ++i)
using namespace std;
const int N = 305;
const int Inf = 0xfffffff;
int n, cnt, s, cur, Max, ans = Inf, head[N], vis[N], dis[N], f[N];

struct Edge {
	int to, next, val;	
}e[N << 1];

inline int read() {
	char ch = gc();
	int ans = 0;
	while (ch > '9' || ch < '0') ch = gc();
	while (ch >= '0' && ch <= '9') ans = (ans << 1) + (ans << 3) + ch - '0', ch = gc();
	return ans;	
}

inline void add(int x, int y, int w) {
	e[++cnt].to = y;
	e[cnt].next = head[x];
	e[cnt].val = w; 
	head[x] = cnt;	
}

inline void dfs1(int x, int fa) {
	f[x] = fa;
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		dis[u] = dis[x] + e[i].val;
		if (dis[u] > Max) Max = dis[u], cur = u;
		dfs1(u, x);
	}
}

inline void dfs2(int x, int fa) {
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		if (!vis[u]) dis[u] = dis[x] + e[i].val, ans = max(ans, dis[u]);
		dfs2(u, x);
	}
}

int main() {
	n = read(), s = read();
	int x, y, w;
	rep(i, 2, n) x = read(), y = read(), w = read(), add(x, y, w), add(y, x, w);
	dfs1(1, 0);
	int l = cur;
	rep(i, 1, n) dis[i] = 0;
	Max = 0, dfs1(l, 0);
	int r = cur; 
	for (int i = r, j = r; i; i = f[i]) {
		vis[i] = 1;
		while (dis[j] - dis[i] > s) j = f[j];
		Max = max(dis[i], dis[r] - dis[j]);
		ans = min(ans, Max);
	}
	rep(i, 1, n) dis[i] = 0;
	dfs2(r, 0);
	printf("%d
", ans);
	return 0;
}