[題目鏈接]

https://www.lydsy.com/JudgeOnline/problem.php?id=4551

[算法]

樹鏈剖分

時間復雜度 : O(QlogN)

[代碼]

#include<bits/stdc++.h>
using namespace std;
#define MAXN 100010

struct Node
{
        int l , r;
        bool flg;
} Tree[MAXN << 2];
struct edge
{
        int to , nxt;
} e[MAXN 
 
<< 1];

int n , q , timer , tot;
int size[MAXN],top[MAXN],dfn[MAXN],head[MAXN],son[MAXN],father[MAXN],rev[MAXN];

template <typename T> inline void chkmax(T &x,T y) { x = max(x,y); }
template <typename T> inline void chkmin(T &x,T y) { x = min(x,y); }
template <typename T> inline void
 
 read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == ‘-‘) f = -f;
    for (; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + c - ‘0‘;
    x *= f;
}
inline void addedge(int u,int v)
{    
        tot++;
        e[tot] = (edge){v,head[u]};
        head[u] 
 
= tot;
}
inline void dfs1(int u)
{
        size[u] = 1;
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (v == father[u]) continue;
                father[v] = u;
                dfs1(v);
                size[u] += size[v];
                if (size[v] > size[son[u]]) son[u] = v;        
        }        
}
inline void dfs2(int u,int tp)
{
        dfn[u] = ++timer;
        rev[timer] = u;
        top[u] = tp;
        if (son[u]) dfs2(son[u],tp);
        for (int i = head[u]; i; i = e[i].nxt)
        {
                int v = e[i].to;
                if (v != son[u] && v != father[u]) dfs2(v,v);        
        }        
}
inline void update(int index)
{
        Tree[index].flg = Tree[index << 1].flg | Tree[index << 1 | 1].flg;
}
inline void build(int index,int l,int r)
{
        Tree[index].l = l; 
        Tree[index].r = r;
        Tree[index].flg = false;
        if (l == r)
        {
                if (l == 1) Tree[index].flg = true;
                return;
        }
        int mid = (l + r) >> 1;
        build(index << 1,l,mid);
        build(index << 1 | 1,mid + 1,r);
        update(index);
}
inline void modify(int index,int pos)
{
        if (Tree[index].l == Tree[index].r)
        {
                Tree[index].flg = true;
                return;
        }
        int mid = (Tree[index].l + Tree[index].r) >> 1;
        if (mid >= pos) modify(index << 1,pos);
        else modify(index << 1 | 1,pos);
        update(index);
}
inline int query(int index,int l,int r)
{
        if (!Tree[index].flg) return 0;
        if (Tree[index].l == Tree[index].r) return l;
        int mid = (Tree[index].l + Tree[index].r) >> 1;
        if (mid >= r) return query(index << 1,l,r);
        else if (mid + 1 <= l) return query(index << 1 | 1,l,r);
        else
        {
                int tmp = query(index << 1 | 1,mid + 1,r);
                if (tmp == 0) tmp = query(index << 1,l,mid);        
                return tmp;
        }        
}

int main()
{
        
        scanf("%d%d",&n,&q);
        for (int i = 1; i < n; i++)
        {
                int u , v;
                scanf("%d%d",&u,&v);
                addedge(u,v);
                addedge(v,u);
        }
        dfs1(1);
        dfs2(1,1);
        build(1,1,n);
        while (q--)
        {
                char op[5];
                int x;
                scanf("%s%d",&op,&x);
                if (op[0] == ‘Q‘)
                {
                        int tx = top[x] , pos = 0;
                        while (true)
                        {
                                pos = query(1,dfn[tx],dfn[x]);
                                if (pos != 0) break;
                                x = father[tx]; tx = top[x];
                        }
                        printf("%d\n",rev[pos]);
                } else modify(1,dfn[x]);
        }
            
        return 0;
    
}

[Tjoi2016&Heoi2016] 樹