题目

非常经典的题目
在 \(Bytemountains\) 有 \(n\) 座山峰，每座山峰有他的高度 \(h_i\) 。有些山峰之间有双向道路相连，共 \(m\) 条路径，每条路径有一个困难值，这个值越大表示越难走。
现在有 \(q\) 组询问，每组询问询问从点 \(v\) 开始只经过困难值小于等于 \(x\) 的路径所能到达的山峰中第 \(k\) 高的山峰，如果无解输出 \(-1\)。

解法

非常经典的解法
考虑 \(Kruskal\) 重构树的性质，然后对于一个子树求第 \(k\) 大，主席树就行了

\(Code\)

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
using namespace std;

const int N = 2e5 + 5, INF = 0x3f3f3f3f;
int n, m, q, a[N], L = INF, R = -INF;

inline void read(int &x)
{
	x = 0; int f = 1; char ch = getchar();
	while (ch < '0' || ch > '9') f = (ch == '-' ? -1 : f), ch = getchar();
	while (ch >= '0' && ch <= '9') x = (x<<3)+(x<<1)+ch-'0', ch = getchar();
	x *= f;
}

int h[N], tot;
struct edge{int to, nxt;}e[N];
inline void add(int x, int y){e[++tot] = edge{y, h[x]}, h[x] = tot;}

int fa[N];
int find(int x){return (fa[x] == x ? x : (fa[x] = find(fa[x])));}

int val[N], num;
struct Edge{int u, v, w;}E[3 * N];
inline bool cmp(Edge x, Edge y){return x.w < y.w;}
void Kruskal()
{
	sort(E + 1, E + m + 1, cmp);
	for(register int i = 1; i < 2 * n; i++) fa[i] = i;
	num = n;
	for(register int i = 1; i <= m; i++)
	{
		int tx = find(E[i].u), ty = find(E[i].v);
		if (tx ^ ty)
		{
			fa[tx] = fa[ty] = ++num, val[num] = E[i].w;
			add(num, tx), add(num, ty);
			if (num == 2 * n - 1) break;
		}
	}
}

int siz[N], dfn[N], rev[N], ed[N], dfc, f[N][20];
void dfs(int x)
{
	if (x <= n) siz[x] = 1, dfn[x] = ++dfc, rev[dfc] = x;
	for(register int i = 1; i <= 18; i++)
	if (f[x][i - 1]) f[x][i] = f[f[x][i - 1]][i - 1];
	else break;
	for(register int i = h[x]; i; i = e[i].nxt)
	{
		int v = e[i].to;
		if (v == fa[x]) continue;
		fa[v] = x, f[v][0] = x, dfs(v), siz[x] += siz[v];
	}
	ed[x] = dfc;
}
inline int getup(int u, int lim, int k)
{
	for(register int i = 18; i >= 0; --i)
	if (f[u][i] && val[f[u][i]] <= lim) u = f[u][i];
	if (siz[u] < k) return -1;
	return u;
}

int size, rt[N];
struct node{int sum, ls, rs;}seg[N * 32];
void update(int &u, int v, int l, int r, int x)
{
	u = ++size;
	seg[u] = seg[v], ++seg[u].sum;
	if (l == r) return;
	int mid = (l + r) >> 1;
	if (x <= mid) update(seg[u].ls, seg[v].ls, l, mid, x);
	else update(seg[u].rs, seg[v].rs, mid + 1, r, x);
}
int query(int u, int v, int l, int r, int k)
{
	if (l == r) return l;
	int mid = (l + r) >> 1, s = seg[seg[u].rs].sum - seg[seg[v].rs].sum;
	if (k <= s) return query(seg[u].rs, seg[v].rs, mid + 1, r, k);
	return query(seg[u].ls, seg[v].ls, l, mid, k - s);
}

int main()
{
	read(n), read(m), read(q);
	for(register int i = 1; i <= n; i++) read(a[i]), L = min(L, a[i]), R = max(R, a[i]);
	for(register int i = 1; i <= m; i++) read(E[i].u), read(E[i].v), read(E[i].w);
	Kruskal(), memset(fa, 0, sizeof fa), dfs(num);
	for(register int i = 1; i <= n; i++) update(rt[i], rt[i - 1], L, R, a[rev[i]]);
	for(int v, x, k; q; --q)
	{
		read(v), read(x), read(k);
		int up = getup(v, x, k);
		printf("%d\n", (up == -1 ? -1 : query(rt[ed[up]], rt[ed[up] - siz[up]], L, R, k)));
	}
}

原文地址：https://www.cnblogs.com/leiyuanze/p/14332723.html