牛客 在二叉树中找到两个节点的最近公共祖先(再进阶)

题目链接：https://www.nowcoder.com/practice/66a813f33f1e499e816ddca216aa1983?tpId=101&tqId=33246&tPage=1&rp=1&ru=/ta/programmer-code-interview-guide&qru=/ta/programmer-code-interview-guide/question-ranking

题目大意

　　略。

分析

　　ST 表的时间复杂度是 O(N*logN + M)，而此题所用的 dfs + 并查集的复杂度为 O(N + M)，然而最终耗时并没有 ST 快，着实令人费解。

　　此题的详解可参看左神的书。

代码如下

#include <bits/stdc++.h>
using namespace std;
 
#define INIT() ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);
#define Rep(i,n) for (int i = 0; i < (int)(n); ++i)
#define For(i,s,t) for (int i = (int)(s); i <= (int)(t); ++i)
#define rFor(i,t,s) for (int i = (int)(t); i >= (int)(s); --i)
#define ForLL(i, s, t) for (LL i = LL(s); i <= LL(t); ++i)
#define rForLL(i, t, s) for (LL i = LL(t); i >= LL(s); --i)
#define foreach(i,c) for (__typeof(c.begin()) i = c.begin(); i != c.end(); ++i)
#define rforeach(i,c) for (__typeof(c.rbegin()) i = c.rbegin(); i != c.rend(); ++i)
 
#define pr(x) cout << #x << " = " << x << "  "
#define prln(x) cout << #x << " = " << x << endl
 
#define LOWBIT(x) ((x)&(-x))
 
#define ALL(x) x.begin(),x.end()
#define INS(x) inserter(x,x.begin())
#define UNIQUE(x) x.erase(unique(x.begin(), x.end()), x.end())
#define REMOVE(x, c) x.erase(remove(x.begin(), x.end(), c), x.end()); // 删去 x 中所有 c 
#define TOLOWER(x) transform(x.begin(), x.end(), x.begin(),::tolower);
#define TOUPPER(x) transform(x.begin(), x.end(), x.begin(),::toupper);
 
#define ms0(a) memset(a,0,sizeof(a))
#define msI(a) memset(a,0x3f,sizeof(a))
#define msM(a) memset(a,-1,sizeof(a))

#define MP make_pair
#define PB push_back
#define ft first
#define sd second
 
template<typename T1, typename T2>
istream &operator>>(istream &in, pair<T1, T2> &p) {
    in >> p.first >> p.second;
    return in;
}
 
template<typename T>
istream &operator>>(istream &in, vector<T> &v) {
    for (auto &x: v)
        in >> x;
    return in;
}

template<typename T>
ostream &operator<<(ostream &out, vector<T> &v) {
    Rep(i, v.size()) out << v[i] << " 
"[i == v.size() - 1];
    return out;
}
 
template<typename T1, typename T2>
ostream &operator<<(ostream &out, const std::pair<T1, T2> &p) {
    out << "[" << p.first << ", " << p.second << "]" << "
";
    return out;
}

inline int gc(){
    static const int BUF = 1e7;
    static char buf[BUF], *bg = buf + BUF, *ed = bg;
    
    if(bg == ed) fread(bg = buf, 1, BUF, stdin);
    return *bg++;
} 

inline int ri(){
    int x = 0, f = 1, c = gc();
    for(; c<48||c>57; f = c=='-'?-1:f, c=gc());
    for(; c>47&&c<58; x = x*10 + c - 48, c=gc());
    return x*f;
}

template<class T>
inline string toString(T x) {
    ostringstream sout;
    sout << x;
    return sout.str();
}

inline int toInt(string s) {
    int v;
    istringstream sin(s);
    sin >> v;
    return v;
}

//min <= aim <= max
template<typename T>
inline bool BETWEEN(const T aim, const T min, const T max) {
    return min <= aim && aim <= max;
}

typedef unsigned int uI;
typedef long long LL;
typedef unsigned long long uLL;
typedef vector< int > VI;
typedef vector< bool > VB;
typedef vector< char > VC;
typedef vector< double > VD;
typedef vector< string > VS;
typedef vector< LL > VL;
typedef vector< VI > VVI;
typedef vector< VB > VVB;
typedef vector< VS > VVS;
typedef vector< VL > VVL;
typedef vector< VVI > VVVI;
typedef vector< VVL > VVVL;
typedef pair< int, int > PII;
typedef pair< LL, LL > PLL;
typedef pair< int, string > PIS;
typedef pair< string, int > PSI;
typedef pair< string, string > PSS;
typedef pair< double, double > PDD;
typedef vector< PII > VPII;
typedef vector< PLL > VPLL;
typedef vector< VPII > VVPII;
typedef vector< VPLL > VVPLL;
typedef vector< VS > VVS;
typedef map< int, int > MII;
typedef unordered_map< int, int > uMII;
typedef map< LL, LL > MLL;
typedef map< string, int > MSI;
typedef map< int, string > MIS;
typedef set< int > SI;
typedef stack< int > SKI;
typedef deque< int > DQI;
typedef queue< int > QI;
typedef priority_queue< int > PQIMax;
typedef priority_queue< int, VI, greater< int > > PQIMin;
const double EPS = 1e-8;
const LL inf = 0x7fffffff;
const LL infLL = 0x7fffffffffffffffLL;
const LL mod = 1e9 + 7;
const int maxN = 1e5 + 7;
const LL ONE = 1;
const LL evenBits = 0xaaaaaaaaaaaaaaaa;
const LL oddBits = 0x5555555555555555;

struct TreeNode {
    int lch = 0, rch = 0, val = 0, fa = 0, level = 0;
};

int N, M, root, o1, o2, ans[maxN];
TreeNode tree[maxN];

unordered_map< int, VI > queryMap; // key 为节点号，value 为询问中与key有关的其他节点号 
unordered_map< int, VI > posMap; // key 为节点号，value 为询问中与key有关答案填写位置 

int dsu[maxN];

inline int Find(int x) {
    while (x != dsu[x]) x = dsu[x] = dsu[dsu[x]];
    return x;
} 

inline void Union(int x, int y){
    dsu[y] = x;
}

inline void dfs(int rt) {
    dsu[rt] = rt;
    
    if(queryMap.find(rt) != queryMap.end()) {
        VI &VQ = queryMap[rt], &VP = posMap[rt];
        
        Rep(i, VQ.size()) {
            if(dsu[VQ[i]]) {
                ans[VP[i]] = Find(VQ[i]);
            }
        }
        
        queryMap.erase(rt);
        posMap.erase(rt);
    }
    
    if(tree[rt].lch) {
        dfs(tree[rt].lch);
        Union(rt, tree[rt].lch);
    }
    
    if(tree[rt].rch) {
        dfs(tree[rt].rch);
        Union(rt, tree[rt].rch);
    }
}

int main(){
    //freopen("MyOutput.txt","w",stdout);
    //freopen("input.txt","r",stdin);
    //INIT();
    scanf("%d%d", &N, &root);
    Rep(i, N) {
        int fa, lch, rch;
        scanf("%d%d%d", &fa, &lch, &rch);
        
        tree[fa].lch = lch;
        tree[fa].rch = rch;
    }
    
    scanf("%d", &M);
    Rep(i, M) {
        scanf("%d%d", &o1, &o2);
        queryMap[o1].PB(o2);
        queryMap[o2].PB(o1);
        posMap[o1].PB(i);
        posMap[o2].PB(i);
    }
    dfs(root);
    
    Rep(i, M) printf("%d
", ans[i]);
    return 0;
}