[SDOI 2013]森林

Description

题库链接

给你 \(n\) 个节点，初始 \(m\) 条边， \(t\) 组操作。让你支持：

询问树上路径点权 \(K\) 小；
支持加边操作。

强制在线，所有状态保证是一个树（森林）。

\(1\leq n,m,t\leq 8\cdot 10^4\)

Solution

分别考虑这道题的两个子问题，显然一个是主席树的板子，另一个是 \(LCT\) 的板子。

这两者不好结合，干脆丢掉一个。

查询 \(K\) 小，暴力的话显然是不可行的。但动态连边我们是可以用启发式合并将复杂度均摊成 \(log\) 。

总复杂度 \(O(nlog_2^2 n)\) ，是可行的。

Code

//It is made by Awson on 2018.3.15

#include <bits/stdc++.h>

#define LL long long

#define dob complex<double>

#define Abs(a) ((a) < 0 ? (-(a)) : (a))

#define Max(a, b) ((a) > (b) ? (a) : (b))

#define Min(a, b) ((a) < (b) ? (a) : (b))

#define Swap(a, b) ((a) ^= (b), (b) ^= (a), (a) ^= (b))

#define writeln(x) (write(x), putchar('\n'))

#define lowbit(x) ((x)&(-(x)))

using namespace std;

const int N = 8e4;

void read(int &x) {

    char ch; bool flag = 0;

    for (ch = getchar(); !isdigit(ch) && ((flag |= (ch == '-')) || 1); ch = getchar());

    for (x = 0; isdigit(ch); x = (x<<1)+(x<<3)+ch-48, ch = getchar());

    x *= 1-2*flag;

}

void print(LL x) {if (x > 9) print(x/10); putchar(x%10+48); }

void write(LL x) {if (x < 0) putchar('-'); print(Abs(x)); }

int n, lim, m, t, a[N+5], b[N+5], tot, u, v, k, fa[N+5], pa[N+5], size[N+5], dep[N+5], f[N+5][20];

struct tt {int to, next; }edge[(N<<1)+5];

int path[N+5], top, last; char ch[5];

struct Segment_tree {

    int root[N+5], ch[N*120+5][2], key[N*120+5], pos;

    int cpynode(int o) {++pos, ch[pos][0] = ch[o][0], ch[pos][1] = ch[o][1], key[pos] = key[o]; return pos; }

    void insert(int &o, int l, int r, int loc) {

    o = cpynode(o); ++key[o];

    if (l == r) return; int mid = (l+r)>>1;

    if (loc <= mid) insert(ch[o][0], l, mid, loc); else insert(ch[o][1], mid+1, r, loc);

    }

    int query(int a, int b, int c, int d, int l, int r, int k) {

    if (l == r) return l; int mid = (l+r)>>1;

    int tmp = key[ch[a][0]]+key[ch[b][0]]-key[ch[c][0]]-key[ch[d][0]];

    if (tmp >= k) return query(ch[a][0], ch[b][0], ch[c][0], ch[d][0], l, mid, k);

    else return query(ch[a][1], ch[b][1], ch[c][1], ch[d][1], mid+1, r, k-tmp);

    }

}T;

void clear(int o) {

    dep[o] = 0;

    for (int i = path[o]; i; i = edge[i].next)

    if (dep[edge[i].to]) clear(edge[i].to);

}

void dfs(int o, int pare, int father, int depth) {

    T.root[o] = T.root[father]; T.insert(T.root[o], 1, tot, lower_bound(b+1, b+tot+1, a[o])-b);

    dep[o] = depth, pa[o] = f[o][0] = father; ++size[pare];

    for (int i = 1; i <= lim; i++) f[o][i] = f[f[o][i-1]][i-1];

    for (int i = path[o]; i; i = edge[i].next)

    if (!dep[edge[i].to]) dfs(edge[i].to, pare, o, depth+1);

}

int query(int u, int v) {

    if (dep[u] < dep[v]) Swap(u, v);

    for (int i = lim; i >= 0; i--) if (dep[f[u][i]] >= dep[v]) u = f[u][i];

    if (u == v) return u;

    for (int i = lim; i >= 0; i--) if (f[u][i] != f[v][i]) u = f[u][i], v = f[v][i];

    return f[u][0];

}

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

void add(int u, int v) {edge[++top].to = v, edge[top].next = path[u], path[u] = top; }

void work() {

    int testcase; read(testcase);

    read(n), read(m), read(t); lim = log(n)/log(2);

    for (int i = 1; i <= n; i++) read(a[i]), b[i] = a[i];

    sort(b+1, b+n+1);

    tot = unique(b+1, b+n+1)-b-1;

    for (int i = 1; i <= m; i++) read(u), read(v), add(u, v), add(v, u), fa[find(u)] = find(v);

    for (int i = 1; i <= n; i++) if (fa[i] == 0) dfs(i, i, 0, 1);

    while (t--) {

    scanf("%s", ch); read(u), u ^= last, read(v), v ^= last;

    if (ch[0] == 'Q') read(k), k ^= last, m = query(u, v), writeln(last = b[T.query(T.root[u], T.root[v], T.root[m], T.root[pa[m]], 1, tot, k)]);

    else {

        if (size[find(u)] < size[find(v)]) Swap(u, v); clear(v);

        add(u, v), add(v, u); fa[find(v)] = find(u);

        dfs(v, find(u), u, dep[u]+1);

    }

    }

}

int main() {

    work(); return 0;

}

巴特西

[SDOI 2013]森林

Description

Solution

Code

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