• [Luogu] 仓鼠找sugar


    https://www.luogu.org/problemnew/show/3398

    树剖练习题,两个懒标记,搜索时序为全局懒标记

    #include <bits/stdc++.h>
    
    using namespace std;
    const int N = 1e5 + 10;
    
    #define gc getchar()
    #define lson jd << 1
    #define rson jd << 1 | 1
    
    struct Node_1{
        int v, nxt;
    }G[N << 1];
    struct Node_2{
        int fa, son, siz, deep, topp, tree;
    }P[N];
    struct Node_3{
        int l, r, f, ff;
    }T[N << 2];
    int n, Ti, head[N], now = 1, tim, imp_clock;
    bool ans, answer;
    
    inline int read(){
        int x = 0; char c = gc;
        while(c < '0' || c > '9') c = gc;
        while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = gc;
        return x;
    }
    
    inline void add(int u, int v){
        G[now].v = v;
        G[now].nxt = head[u];
        head[u] = now ++;
    }
    
    void dfs_find_son(int u, int fa, int dep){
        P[u].fa = fa;
        P[u].deep = dep;
        P[u].siz = 1;
        for(int i = head[u]; ~ i; i = G[i].nxt){
            int v = G[i].v;
            if(v != fa){
                dfs_find_son(v, u, dep + 1);
                P[u].siz += P[v].siz;
                if(P[v].siz > P[P[u].son].siz) P[u].son = v;
            }
        }
    }
    
    void dfs_to_un(int u, int tp){
        P[u].topp = tp;
        P[u].tree = ++ tim;
        if(!P[u].son) return ;
        dfs_to_un(P[u].son, tp);
        for(int i = head[u]; ~ i; i = G[i].nxt){
            int v = G[i].v;
            if(v != P[u].fa && v != P[u].son) dfs_to_un(v, v);
        }
    }
    
    void Sec_G(int l, int r, int jd, int x, int y, int g){
        if(x <= l && r <= y) {
            T[jd].f = g;
            T[jd].ff = g;
            return ;
        }
        int mid = (l + r) >> 1;
        if(x <= mid) Sec_G(l, mid, lson, x, y, g);
        if(y > mid) Sec_G(mid + 1, r, rson, x, y, g);
        if(T[lson].f == g || T[rson].f == g) T[jd].f = g;
        
    }
    
    void Sec_G_imp(int x, int y){
        ++ imp_clock;
        int tp1 = P[x].topp, tp2 = P[y].topp;
        while(tp1 != tp2){
            if(P[tp1].deep < P[tp2].deep) swap(x, y), swap(tp1, tp2);
            Sec_G(1, n, 1, P[tp1].tree, P[x].tree, imp_clock);
            x = P[tp1].fa;
            tp1 = P[x].topp;
        }
        if(P[x].deep > P[y].deep) swap(x, y);
        Sec_G(1, n, 1, P[x].tree, P[y].tree, imp_clock);
    }
    
    void Sec_pd(int l, int r, int jd, int x, int y){
        if(x <= l && r <= y){
            if(T[jd].f == imp_clock) ans = 1;
            return ;
        }
        if(T[jd].ff == imp_clock && T[jd].f == imp_clock){
            T[lson].f = T[rson].f = T[jd].f;
            T[lson].ff = T[rson].ff = imp_clock;    
        } 
        int mid = (l + r) >> 1;
        if(x <= mid) Sec_pd(l, mid, lson, x, y);
        if(y > mid) Sec_pd(mid + 1, r, rson, x, y);
    }
    
    bool Sec_pd_imp(int x, int y){
        int tp1 = P[x].topp, tp2 = P[y].topp;
        while(tp1 != tp2){
            if(P[tp1].deep < P[tp2].deep) swap(x, y), swap(tp1, tp2);
            ans = 0;
            Sec_pd(1, n, 1, P[tp1].tree, P[x].tree);
            if(ans) return 1;
            x = P[tp1].fa;
            tp1 = P[x].topp;
        }
        ans = 0;
        if(P[x].deep > P[y].deep) swap(x, y);
        Sec_pd(1, n, 1, P[x].tree, P[y].tree);
        if(ans) return 1;
        return 0;
    }
    
    inline void work(int x_1, int y_1, int x_2, int y_2){
        Sec_G_imp(x_1, y_1);
        answer = Sec_pd_imp(x_2, y_2);
        if(answer) puts("Y");
        else puts("N");
        return ;
    }
    
    int main()
    {
        n = read(); Ti = 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_find_son(1, 0, 1); 
        dfs_to_un(1, 1);
        while(Ti --){
            int x_1 = read(), y_1 = read(), x_2 = read(), y_2 = read();
            work(x_1, y_1, x_2, y_2);
        }
        return 0;
    }
    /*
    5 5
    2 5
    4 2
    1 3
    1 4
    5 1 5 1
    2 2 1 4
    4 1 3 4
    3 1 1 5
    3 5 1 4
    */
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  • 原文地址:https://www.cnblogs.com/shandongs1/p/8011223.html
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