[Luogu] U18430 萌萌的大河

https://www.luogu.org/problemnew/show/U18430

思路比较好想

树链剖分

对于1操作

　　只需将以该点为根的子树打标记，将所有数存入数组排序

　　每次进行1操作时，判断若该点已经被标记过，则操作无效

对于操作2

　　在原树中求出lca，若lca被标记过，lca就是询问的两个点中

　　编号较小的那个，lca值不改变

对于操作3

　　若该点打过标记，输出该点与根的这条链上被打过标记的

　　深度最小节点编号a与该节点编号的差+a的深度

 

 

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cmath>

using namespace std;
const int N = 1e6 + 10;

#define yxy getchar()
#define lson jd << 1
#define rson jd << 1 | 1

int n, Q, now = 1, tim, w1, w2, w3;
int head[N], Num[N], deep[N], fa[N], lnum[N], rnum[N], son[N], topp[N], siz[N], tree[N], g_min[N << 2];
struct Node{int u, v, nxt;} G[N << 1];

inline int read(){
    int x = 0; char c = yxy;
    while(c < '0' || c > '9') c = yxy;
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = yxy;
    return x;
}

inline void add(int u, int v){
    G[now].u = u; G[now].v = v; G[now].nxt = head[u]; head[u] = now ++;
}

void dfs_1(int u, int father, int dep){
    fa[u] = father;
    deep[u] = dep;
    siz[u] = 1;
    for(int i = head[u]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(v != father){
            dfs_1(v, u, dep + 1);
            siz[u] += siz[v];
            if(siz[v] > siz[son[u]]) son[u] = v;
        }
    }
}

void dfs_2(int u, int tp){
    topp[u] = tp;
    lnum[u] = ++ tim;
    tree[u] = tim;
    Num[tim] = u;
    if(!son[u]) {rnum[u] = tim; return ;}
    dfs_2(son[u], tp);
    for(int i = head[u]; ~ i; i = G[i].nxt){
        int v = G[i].v;
        if(v != fa[u] && v != son[u]) dfs_2(v, v);
    }
    rnum[u] = tim;
}

void Poi_A(int l, int r, int jd, int x){
    if(l == r) {
        if(g_min[jd]) w1 = 1;
        return ;
    }
    if(g_min[jd]) {
        g_min[lson] = g_min[jd];
        g_min[rson] = g_min[jd];
    }
    int mid = (l + r) >> 1;
    if(x <= mid) Poi_A(l, mid, lson, x);
    else Poi_A(mid + 1, r, rson, x);
}

void Sec_G(int l, int r, int jd, int x, int y, int yg){
    if(x <= l && r <= y){
        g_min[jd] = yg;
        return ;
    }
    int mid = (l + r) >> 1;
    if(x <= mid) Sec_G(l, mid, lson, x, y, yg);
    if(y > mid)  Sec_G(mid + 1, r, rson, x, y, yg);
}

int calc_lca(int x, int y){
    int tp1 = topp[x], tp2 = topp[y];
    while(tp1 != tp2){
        if(deep[tp1] < deep[tp2]) swap(tp1, tp2), swap(x, y);
        x = fa[tp1];
        tp1 = topp[x];
    }
    return deep[x] > deep[y] ? y : x;
}

void Sec_A(int l, int r, int jd, int x, int y){
    if(x <= l && r <= y){
        w2 = g_min[jd];
        return ;
    }
    if(g_min[jd]) {
        g_min[lson] = g_min[jd];
        g_min[rson] = g_min[jd];
    }
    int mid = (l + r) >> 1;
    if(x <= mid) Sec_A(l, mid, lson, x, y);
    if(y > mid)  Sec_A(mid + 1, r, rson, x, y);
}

int get_answer(int l, int r, int x){
    int L = l, R = r, ret;
    while(L <= R){
        int mid = (L + R) >> 1;
        if(Num[mid] <= x) L = mid + 1, ret = mid;
        else R = mid - 1;
    }
    return ret - l;
}

int main()
{
    n = read(); Q = read();
    for(int i = 1; i <= n; i ++) head[i] = -1;
    for(int i = 1; i < n; i ++) {
        int u = read(), v = read();
        add(u, v); add(v, u);
    }
    dfs_1(1, 0, 1);
    dfs_2(1, 1);
    while(Q --){
        int opt = read();
        if(opt == 1){
            int x = read();
            if(!son[x]) continue ;
            w1 = 0;
            Poi_A(1, n, 1, tree[x]);
            if(w1) continue ;
            Sec_G(1, n, 1, lnum[x], rnum[x], x);
            sort(Num + lnum[x] + 1, Num + rnum[x] + 1); 
        }
        else if(opt == 2){
            int x = read(), y = read();
            int Lca_now = calc_lca(x, y);
            w1 = 0;
            Poi_A(1, n, 1, tree[Lca_now]);
            if(w1) printf("%d\n", min(x, y));
            else printf("%d\n", Lca_now);
        }
        else {
            int x = read();
            w1 = 0;
            Poi_A(1, n, 1, tree[x]);
            if(!w1) printf("%d\n", deep[x]);
            else {
                Sec_A(1, n, 1, lnum[x], rnum[x]);
                printf("%d\n", get_answer(lnum[w2], rnum[w2], x) + deep[w2]);
            }
        }
    }
    return 0;
}