[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\n", 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[] = {, , -, , , , -, -};
 const int dy[] = {-, , , , , -, , - };
 const int maxn = 4e5 + ;
 const int md = 1e9 + ;
 const int inf = 1e9 + ;
 const LL inf_L = 1e18 + ;
 const double pi = acos(-1.0);
 const double eps = 1e-;

 template<class T>T gcd(T a, T b){return b==?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][];
     int t[maxn];
     void Init_min(int a[], int n) {
         rep_up0(i, n) d[i][] = a[i];
         for (int j = ; ( << j) <= n; j++) {
             for (int i = ; i + ( << j) -  < n; i++) {
                 d[i][j] = min(d[i][j - ], d[i + ( << (j - ))][j - ]);
             }
         }
         int p = -;
         rep_up1(i, n) {
             if ((i & (i - )) == ) p++;
             t[i] = p;
         }
     }
     void Init_max(int a[], int n) {
         rep_up0(i, n) d[i][] = a[i];
         for (int j = ; ( << j) <= n; j++) {
             for (int i = ; i + ( << j) -  < n; i++) {
                 d[i][j] = max(d[i][j - ], d[i + ( << (j - ))][j - ]);
             }
         }
         int p = -;
         rep_up1(i, n) {
             if ((i & (i - ) == )) p++;
             t[i] = p;
         }
     }
     int RMQ_min(int L, int R) {
         int p = t[R - L + ];
         return min(d[L][p], d[R - ( << p) + ][p]);
     }
     int RMQ_max(int L, int R) {
         int p = t[R - L + ];
         return max(d[L][p], d[R - ( << p) + ][p]);
     }
 };

 SparseTable st;

 struct Graph {
     vector<vector<int> > G;
     void clear() { G.clear(); }
     void resize(int n) { G.resize(n + ); }
     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 - ) {
             int u, v;
             sint2(u, v);
             G.add(u, v);
             G.add(v, u);
         }
         dfs_clock = ;
         dfs(, );
         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 ;
 }