https://www.patest.cn/contests/gplt/L3-016
题意：中文。
思路：暴力构造，暴力查询就好了。只不过操作很多，很麻烦。比赛的时候再给我10分钟就打完了，手速太慢好可惜。
1 #include <bits/stdc++.h> 2 using namespace std; 3 #define N 100010 4 typedef long long LL; 5 struct Node { 6 int num; 7 Node *l, *r, *fa; 8 Node (int n) { 9 l = r = fa = NULL; 10 num = n; 11 } 12 } ; 13 Node *root; 14 int flag, d; 15 char s[5][15]; 16 17 void Insert(Node *now, Node *f, int x, int kind) { 18 if(now == NULL) { 19 now = new Node(x); 20 now->fa = f; 21 if(kind) f->r = now; 22 else f->l = now; 23 return ; 24 } 25 if(x > now->num) Insert(now->r, now, x, 1); 26 else Insert(now->l, now, x, 0); 27 } 28 29 Node *Find(int x) { 30 Node *now = root; d = 0; 31 while(now != NULL && now->num != x) { 32 if(x > now->num) now = now->r; 33 else now = now->l; 34 d++; 35 } 36 return now; 37 } 38 39 void Solve1(int x) { 40 if(root && root->num == x) flag = 1; 41 } 42 43 void Solve2(int x, int y) { 44 Node *now1 = Find(x), *now2 = Find(y); 45 if(now1 && now2 && now1->fa == now2->fa) flag = 1; 46 } 47 48 void Solve3(int x, int y) { 49 Node *now = Find(y); 50 if(now && now->fa && now->fa->num == x) flag = 1; 51 } 52 53 void Solve4(int x, int y) { 54 Node *now = Find(y); 55 if(now && now->l && now->l->num == x) flag = 1; 56 } 57 58 void Solve5(int x, int y) { 59 Node *now = Find(y); 60 if(now && now->r && now->r->num == x) flag = 1; 61 } 62 63 void Solve6(int x, int y) { 64 Node *now1 = Find(x); 65 int d1 = d; 66 Node *now2 = Find(y); 67 int d2 = d; 68 if(now1 && now2 && d1 == d2) flag = 1; 69 } 70 71 int main() { 72 int n, a, b; scanf("%d", &n); 73 if(n) scanf("%d", &a), root = new Node(a); 74 for(int i = 1; i < n; i++) scanf("%d", &a), Insert(root, NULL, a, 0); 75 int q; scanf("%d", &q); 76 while(q--) { 77 flag = 0; 78 scanf("%d", &a); 79 scanf("%s", s[0]); 80 if(s[0][0] == 'a') { 81 scanf("%d", &b); 82 scanf("%s %s", s[1], s[2]); 83 if(s[2][0] == 's') { 84 Solve2(a, b); 85 } else { 86 Solve6(a, b); 87 scanf("%s %s %s", s[3], s[3], s[3]); 88 } 89 } else { 90 scanf("%s %s", s[1], s[2]); 91 if(s[2][1] == 'o') { 92 Solve1(a); 93 } else if(s[2][1] == 'a') { 94 scanf("%s %d", s[3], &b), Solve3(a, b); 95 } else if(s[2][1] == 'e') { 96 scanf("%s %s %d", s[3], s[4], &b), Solve4(a, b); 97 } else { 98 scanf("%s %s %d", s[3], s[4], &b), Solve5(a, b); 99 } 100 } 101 if(flag) puts("Yes"); 102 else puts("No"); 103 } 104 return 0; 105 }