题目传送门:CF150E。
题意:
给定一棵 \(n\)(\(2\le n\le 10^5\))个点的树,边有边权。
定义一条路径的权值为路径经过的边的边权的中位数,若经过偶数条边则取两个中位数中较大的那个。
求长度介于 \(l\) 到 \(r\)(\(1\le l\le r<n\))之间的路径的最大权值,并输出这个路径的两端点。
题解:
看到中位数的定义,首先想到二分答案,假设二分的值为 \(\mathrm{mid}\),将边权 \(\ge\mathrm{mid}\) 的边看作 \(+1\),将边权 \(<\mathrm{mid}\) 的边看作 \(-1\),则一条路径的权值大于等于 \(\mathrm{mid}\) 当且仅当其经过的边的和大于等于 \(0\)。
二分了一个值后,考虑使用点分治统计路径。
合并子树时相当于查询一个滑动窗口内的最大值,用单调队列维护即可。
对当前分治块内统计时要注意需要先处理较小的子树以保证复杂度。
下面是代码,时间复杂度 \(\mathcal{O}(n\log^2 n)\)。
#include <cstdio>
#include <vector>
#include <algorithm>
const int Inf = 0x3f3f3f3f;
const int MN = 100005;
int N, L, R, Ans = -1, AnsU, AnsV;
int uv[MN], w[MN];
std::vector<int> G[MN];
int dw[MN], M;
int vis[MN], siz[MN], tsiz, rsiz, Root;
void GetRoot(int u, int fz) {
siz[u] = 1;
int nsiz = 0;
for (auto i : G[u]) {
int v = uv[i] ^ u;
if (v == fz || vis[v]) continue;
GetRoot(v, u), siz[u] += siz[v];
if (nsiz < siz[v]) nsiz = siz[v];
}
if (nsiz < tsiz - siz[u]) nsiz = tsiz - siz[u];
if (rsiz > nsiz) rsiz = nsiz, Root = u;
}
int stk[MN], tp, _U;
inline bool cmp(int i, int j) {
return siz[uv[i] ^ _U] < siz[uv[j] ^ _U];
}
int seq[MN], sequ[MN], odep, tmp[MN], tmpu[MN], ndep;
void DFS(int u, int fz, int d, int x, int y) {
if (tmp[d] < x) tmp[d] = x, tmpu[d] = u;
if (ndep < d) ndep = d;
for (auto i : G[u]) {
int v = uv[i] ^ u;
if (v == fz || vis[v]) continue;
DFS(v, u, d + 1, x + (w[i] >= y ? 1 : -1), y);
}
}
int ucal, vcal;
bool Calc(int u, int x) {
static int que[MN];
seq[odep = 0] = 0, sequ[0] = u;
for (int i = 1; i <= tp; ++i) {
int v = uv[stk[i]] ^ u;
for (int j = 1; j <= siz[v]; ++j) tmp[j] = -Inf;
ndep = 0, DFS(v, u, 1, w[stk[i]] >= x ? 1 : -1, x);
int l = 1, r = 0, lb = odep, rb = odep + 1;
for (int j = 1; j <= ndep; ++j) {
while (rb > 0 && rb > L - j) {
--rb;
while (l <= r && seq[que[r]] < seq[rb]) --r;
que[++r] = rb;
}
while (lb >= 0 && lb > R - j) {
--lb;
while (l <= r && que[l] > lb) ++l;
}
if (l <= r && seq[que[l]] + tmp[j] >= 0) {
ucal = sequ[que[l]], vcal = tmpu[j];
return 1;
}
}
while (odep < ndep) seq[++odep] = -Inf;
for (int j = 1; j <= ndep; ++j)
if (seq[j] < tmp[j])
seq[j] = tmp[j], sequ[j] = tmpu[j];
}
return 0;
}
void Solve(int u) {
int nsiz = tsiz;
tp = 0;
for (auto i : G[u]) {
int v = uv[i] ^ u;
if (vis[v]) continue;
siz[v] = siz[v] > siz[u] ? nsiz - siz[u] : siz[v];
stk[++tp] = i;
}
_U = u, std::sort(stk + 1, stk + tp + 1, cmp);
int lb = 1, rb = M, mid, ans = 0, ansu = 0, ansv = 0;
while (lb <= rb) {
mid = (lb + rb) >> 1;
if (Calc(u, dw[mid])) {
ans = mid;
ansu = ucal, ansv = vcal;
lb = mid + 1;
}
else rb = mid - 1;
}
if (Ans < dw[ans]) {
Ans = dw[ans];
AnsU = ansu, AnsV = ansv;
}
vis[u] = 1;
for (auto i : G[u]) {
int v = uv[i] ^ u;
if (vis[v]) continue;
rsiz = tsiz = siz[v], GetRoot(v, 0), Solve(Root);
}
}
int main() {
scanf("%d%d%d", &N, &L, &R);
for (int i = 1; i < N; ++i) {
int x, y;
scanf("%d%d%d", &x, &y, &w[i]);
uv[i] = x ^ y;
G[x].push_back(i);
G[y].push_back(i);
dw[i] = w[i];
}
std::sort(dw + 1, dw + N);
M = std::unique(dw + 1, dw + N) - dw - 1;
rsiz = tsiz = N, GetRoot(1, 0), Solve(Root);
printf("%d %d\n", AnsU, AnsV);
return 0;
}