Luogu P4197 Peaks

题目链接 (Click) (Here)

做法：(Kruskal)重构树上跑主席树

构造方法：把每条边拆出来成一个点，点权是原先的边权。每次连边的时候，连的不再是点，而是其原先点所在的联通块。

(Kruskal)重构树的性质：

• 二叉树

• 如果按最小生成树构造，则自顶到底权值递增（堆）

• 叶节点是原先的树上节点

• 两个在原图中可以经过权值(<=k)的边互相抵达的点，在最小生成树上也可以这样互相抵达，在(Kruskal)重构树上同理。

也就是说，假如一个点不能到比它的权值大的点，那么它能抵达的所有点就在它的子树里面。这个题目里我们从询问点跳到权值(<=x)中权值最大的点，其子树中的叶子节点，就是可以询问点可以通过点权(<=x)的点到达的所有点。

#include <bits/stdc++.h>
using namespace std;

const int N = 200010;
const int M = 500010;

int n, m, q, tot, INF, h[N], fa[N], val[N];
struct edge {int nxt, to;} e[N];
struct _edge {int u, v, w;}_e[M];

bool cmp (_edge lhs, _edge rhs) {
    return lhs.w < rhs.w;
}

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

int cnt, head[N];

void add_edge (int u, int v) {
    e[++cnt] = (edge) {head[u], v}; head[u] = cnt;
}

int rt[N];
#define mid ((l + r) >> 1)

struct Segment_Node {
    int sz, ls, rs;
}t[N << 5];

int modify (int _rt, int l, int r, int w) {
    int p = ++rt[0];
    t[p].sz = t[_rt].sz + 1;
    if (l != r) {
        if (w <= mid) {
            t[p].ls = modify (t[_rt].ls, l, mid, w), t[p].rs = t[_rt].rs;
        } else {
            t[p].rs = modify (t[_rt].rs, mid + 1, r, w), t[p].ls = t[_rt].ls;
        }
    } else {
        t[p].ls = t[p].rs = 0;
    }
    return p;
}

int sz[N], id[N], dfn[N], deep[N], fafa[N][20], totsize[N];

int sep[N];

int num (int x) {
    return lower_bound (sep + 1, sep + 1 + sep[0], x) - sep;
}

void dfs (int u, int _fa) {
    sz[u] = head[u] == 0;
	dfn[u] = ++dfn[0];
	totsize[u] = 1;
	id[dfn[0]] = u;
	fafa[u][0] = _fa;
    deep[u] = deep[_fa] + 1;
    rt[u] = modify (rt[id[dfn[u] - 1]], 1, INF, num (h[u])); //主席树维护高度
    for (int i = 1; (1 << i) <= deep[u]; ++i) {
        fafa[u][i] = fafa[fafa[u][i - 1]][i - 1];
    }
    for (int i = head[u]; i; i = e[i].nxt) {
        int v = e[i].to;
        if (v != _fa) {
            dfs (v, u);
            sz[u] += sz[v];
            totsize[u] += totsize[v];
        }
    }
}

int query (int u, int v, int l, int r, int k) {
    while (l < r) {
        int rch = t[t[v].rs].sz - t[t[u].rs].sz;
        if (k <= rch) {
            u = t[u].rs;
            v = t[v].rs;
            l = mid + 1;
        } else {
            u = t[u].ls;
            v = t[v].ls;
            k -= rch;
            r = mid;
        } 
    }
    return l;
}

int read () {
	int s = 0, w = 1, ch = getchar ();
	while ('9' < ch || ch < '0') {
		if (ch == '-') w = -1;
		ch = getchar ();
	}
	while ('0' <= ch && ch <= '9') {
		s = (s << 1) + (s << 3) + ch - '0';
		ch = getchar ();
	}
	return s * w;
} 

int main () {
	memset (val, 0x3f, sizeof (val));
	tot = n = read (), m = read (), q = read ();
    t[0].sz = t[0].ls = t[0].rs = 0, sep[++sep[0]] = 0;
    for (int i = 1; i <= n; ++i) sep[++sep[0]] = h[i] = read ();
    for (int i = 1; i <= m; ++i) {
	    _e[i].u = read ();
	    _e[i].v = read ();
		_e[i].w = read ();
    }
    sort (_e + 1, _e + 1 + m, cmp);
    for (int i = 1; i <= n; ++i) fa[i] = i;
    for (int i = 1; i <= m; ++i) {
        int u = find (_e[i].u);
        int v = find (_e[i].v);
        if (u != v) {
            int T = ++tot;
            val[T] = _e[i].w;
            fa[u] = fa[v] = fa[T] = T;
            add_edge (T, u);
            add_edge (T, v);
        }
    }
    sort (sep + 1, sep + 1 + sep[0]);
    INF = sep[0] = unique (sep + 1, sep + 1 + sep[0]) - sep - 1;
    dfs (tot, 0);
	int lastans = 0;
    for (int i = 1; i <= q; ++i) {
        int u = read () ^ lastans;
		int x = read () ^ lastans;
		int k = read () ^ lastans;
        for (int j = 19; j >= 0; --j) {
			if (val[fafa[u][j]] <= x) {
                u = fafa[u][j];
            }
        }
        if (sz[u] < k) {
            puts ("-1");
			lastans = 0;
        } else {
			lastans = sep[query (rt[id[dfn[u] - 1]], rt[id[dfn[u] + totsize[u] - 1]], 1, INF, k)];
            printf ("%d
", lastans);
        }
    }
}