[csu/coj 1079]树上路径查询 LCA

题意：询问树上从u到v的路径是否经过k

思路：把树dfs转化为有根树后，对于u,v的路径而言，设p为u，v的最近公共祖先，u到v的路径必定是可以看成两条路径的组合，u->p，v->p，这样一来便可以将判断条件转化为(LCA(u,k)=k  || LCA(v,k)=k) && LCA(k,p)=p。由于这个LCA询问里面需要用到中间结果p，所以这种方法用tarjan离线不行，只能用dfs+RMQ。

#pragma comment(linker, "/STACK:10240000,10240000")

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstdlib>
#include <cstring>
#include <map>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <ctime>
#include <cctype>
#include <stack>
#include <set>
#include <bitset>
#include <functional>
#include <numeric>
#include <stdexcept>
#include <utility>

using namespace std;

#define mem0(a) memset(a, 0, sizeof(a))
#define mem_1(a) memset(a, -1, sizeof(a))
#define lson l, m, rt << 1
#define rson m + 1, r, rt << 1 | 1
#define define_m int m = (l + r) >> 1
#define rep_up0(a, b) for (int a = 0; a < (b); a++)
#define rep_up1(a, b) for (int a = 1; a <= (b); a++)
#define rep_down0(a, b) for (int a = b - 1; a >= 0; a--)
#define rep_down1(a, b) for (int a = b; a > 0; a--)
#define all(a) (a).begin(), (a).end()
#define lowbit(x) ((x) & (-(x)))
#define constructInt4(name, a, b, c, d) name(int a = 0, int b = 0, int c = 0, int d = 0): a(a), b(b), c(c), d(d) {}
#define constructInt3(name, a, b, c) name(int a = 0, int b = 0, int c = 0): a(a), b(b), c(c) {}
#define constructInt2(name, a, b) name(int a = 0, int b = 0): a(a), b(b) {}
#define pchr(a) putchar(a)
#define pstr(a) printf("%s", a)
#define sstr(a) scanf("%s", a)
#define sint(a) scanf("%d", &a)
#define sint2(a, b) scanf("%d%d", &a, &b)
#define sint3(a, b, c) scanf("%d%d%d", &a, &b, &c)
#define pint(a) printf("%d
", a)
#define test_print1(a) cout << "var1 = " << a << endl
#define test_print2(a, b) cout << "var1 = " << a << ", var2 = " << b << endl
#define test_print3(a, b, c) cout << "var1 = " << a << ", var2 = " << b << ", var3 = " << c << endl
#define mp(a, b) make_pair(a, b)
#define pb(a) push_back(a)

typedef long long LL;
typedef pair<int, int> pii;
typedef vector<int> vi;

const int dx[8] = {0, 0, -1, 1, 1, 1, -1, -1};
const int dy[8] = {-1, 1, 0, 0, 1, -1, 1, -1 };
const int maxn = 4e5 + 7;
const int md = 1e9 + 7;
const int inf = 1e9 + 7;
const LL inf_L = 1e18 + 7;
const double pi = acos(-1.0);
const double eps = 1e-6;

template<class T>T gcd(T a, T b){return b==0?a:gcd(b,a%b);}
template<class T>bool max_update(T &a,const T &b){if(b>a){a = b; return true;}return false;}
template<class T>bool min_update(T &a,const T &b){if(b<a){a = b; return true;}return false;}
template<class T>T condition(bool f, T a, T b){return f?a:b;}
template<class T>void copy_arr(T a[], T b[], int n){rep_up0(i,n)a[i]=b[i];}
int make_id(int x, int y, int n) { return x * n + y; }

struct SparseTable {
    int d[maxn][22];
    int t[maxn];
    void Init_min(int a[], int n) {
        rep_up0(i, n) d[i][0] = a[i];
        for (int j = 1; (1 << j) <= n; j++) {
            for (int i = 0; i + (1 << j) - 1 < n; i++) {
                d[i][j] = min(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
            }
        }
        int p = -1;
        rep_up1(i, n) {
            if ((i & (i - 1)) == 0) p++;
            t[i] = p;
        }
    }
    void Init_max(int a[], int n) {
        rep_up0(i, n) d[i][0] = a[i];
        for (int j = 1; (1 << j) <= n; j++) {
            for (int i = 0; i + (1 << j) - 1 < n; i++) {
                d[i][j] = max(d[i][j - 1], d[i + (1 << (j - 1))][j - 1]);
            }
        }
        int p = -1;
        rep_up1(i, n) {
            if ((i & (i - 1) == 0)) p++;
            t[i] = p;
        }
    }
    int RMQ_min(int L, int R) {
        int p = t[R - L + 1];
        return min(d[L][p], d[R - (1 << p) + 1][p]);
    }
    int RMQ_max(int L, int R) {
        int p = t[R - L + 1];
        return max(d[L][p], d[R - (1 << p) + 1][p]);
    }
};

SparseTable st;

struct Graph {
    vector<vector<int> > G;
    void clear() { G.clear(); }
    void resize(int n) { G.resize(n + 2); }
    void add(int u, int v) { G[u].push_back(v); }
    vector<int> & operator [] (int u) { return G[u]; }
};
Graph G;

int dfs_clock;
int a[maxn], b[maxn], r[maxn];

void dfs(int u, int fa) {
    r[u] = dfs_clock;
    a[dfs_clock ++] = u;
    int sz = G[u].size();
    rep_up0(i, sz) {
        int v = G[u][i];
        if (fa != v) {
            dfs(v, u);
            a[dfs_clock ++] = u;
        }
    }
}

int LCA(int u, int v) {
    if (r[u] > r[v]) swap(u, v);
    return a[st.RMQ_min(r[u], r[v])];
}

bool chk(int u, int v, int w) {
    int pos = LCA(u, v);
    return (LCA(u, w) == w || LCA(v, w) == w) && LCA(pos, w) == pos;
}

int main() {
    //freopen("in.txt", "r", stdin);
    int n, q;
    while (cin >> n >> q) {
        G.clear();
        G.resize(n);
        rep_up0(i, n - 1) {
            int u, v;
            sint2(u, v);
            G.add(u, v);
            G.add(v, u);
        }
        dfs_clock = 0;
        dfs(1, 0);
        rep_up0(i, dfs_clock) b[i] = r[a[i]];
        st.Init_min(b, dfs_clock);
        rep_up0(i, q) {
            int u, v, w;
            sint3(u, v, w);
            puts(chk(u, v, w)? "YES" : "NO");
        }
        cout << endl;
    }
    return 0;
}