洛谷P5536 【XR-3】核心城市

(Large extbf{Description: } large{有一棵n个节点的树，选择其中的k个相连的节点，求其余所有节点与选择节点之间的距离的最大值的最小值。(1 leq n leq 10^{5})}\)

(Large extbf{Solution: } large{首先一定有一个点在这棵数的直径上，因为如果不在，肯定无法比直径更优。\然后我们可以从直径中点出发，每次往与中点距离最大的点扩展，扩展k次即可。\}\)

(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)
using namespace std;
const int N = 1e5 + 5;
int n, k, cnt, Max, cur, m, head[N], dis[N], f[N], vis[N], d[N], ma[N];

struct Edge {
	int to, next;	
}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) {
	e[++cnt].to = y;
	e[cnt].next = head[x];
	head[x] = cnt;	
}

inline void dfs(int x, int fa) {
	f[x] = fa;
	dis[x] = dis[fa] + 1;
	if (dis[x] > Max) Max = dis[x], cur = x; 
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		dfs(u, x);
	}
}

inline bool cmp(int x, int y) {
	return x > y;
}

inline void dfs2(int x, int fa) {
	dis[x] = ma[x] = dis[fa] + 1;
	for (int i = head[x]; i ; i = e[i].next) {
		int u = e[i].to;
		if (u == fa) continue;
		dfs2(u, x);
		ma[x] = max(ma[x], ma[u]);
	}
}

int main() {
	n = read(), k = read();
	int x, y;
	rep(i, 2, n) x = read(), y = read(), add(x, y), add(y, x);
	dfs(1, 0);
	int l = cur; Max = 0, dfs(l, 0);
	int r = cur, pre = 0;
	for (int i = r; i ; i = f[i]) {
		if (pre >= dis[r] / 2) { m = i; break; }
		++pre;
	}
	dfs2(m, 0); 
	rep(i, 1, n) d[i] = max(d[i], ma[i] - dis[i]);
	sort(d + 1, d + 1 + n, cmp);
	printf("%d
", d[k + 1] + 1);
	return 0;
}