[Luogu 3398] 仓鼠找sugar

又是 LCA…

前两天死活写不过的一个题今天终于顺手切了。

思路嘛参考了一楼题解。

就是说，对于 a, b, c, d 四个点，

令 x = LCA(a, b), y = LCA(c, d)，

两条路径有交叉，当且仅当 c, d 至少一个在 x 的子树下，且 a, b 至少一个在 y 的子树下。

由于我是 HLD 求的 LCA，第一遍 DFS 时顺手把子树大小求了，后边判断在不在一棵子属下的时候就可以很方便了。

就这样。

#include <algorithm>
#include <cstdio>const int MAXN = 100010; int n, q; struct Graph
{struct Edge{int to; Edge *next; Edge(int to, Edge* next): to(to), next(next) {}~Edge(void){if(next != NULL)delete next; }}*head[MAXN]; Graph(int n){std :: fill(head + 1, head + n + 1, (Edge*)NULL); }~Graph(void){for(int i = 1; i <= n; ++i)delete head[i]; }void AddEdges(int u, int v){head[u] = new Edge(v, head[u]); head[v] = new Edge(u, head[v]); }
}*G; namespace HLD
{int num; struct Node{int depth, father, son, top, size, DFN; }s[MAXN]; void DFS1(int u, int k){s[u].depth = k; s[u].size = 1; int v; for(Graph :: Edge *i = G -> head[u]; i != NULL; i = i -> next)if(!s[v = i -> to].size){DFS1(v, k + 1); s[u].size += s[v].size; s[v].father = u; if(s[v].size > s[s[u].son].size)s[u].son = v; }}void DFS2(int u, int top){s[u].top = top; s[u].DFN = ++num; if(s[u].son)DFS2(s[u].son, top); int v; for(Graph :: Edge *i = G -> head[u]; i != NULL; i = i -> next)if(!s[v = i -> to].DFN)DFS2(v, v); }void Init(void){DFS1(1, 1); DFS2(1, 1); }int LCA(int x, int y){int a, b; while((a = s[x].top) ^ (b = s[y].top))if(s[a].depth > s[b].depth)x = s[a].father; elsey = s[b].father; return s[x].depth < s[y].depth ? x : y; }bool Range(int x, int y){return s[x].DFN <= s[y].DFN && s[y].DFN < s[x].DFN + s[x].size; }bool Query(int a, int b, int c, int d){int x = LCA(a, b), y = LCA(c, d); return (Range(x, c) || Range(x, d)) && (Range(y, a) || Range(y, b)); }
}int main(void)
{scanf("%d %d", &n, &q); G = new Graph(n); for(int i = 1, u, v; i < n; ++i){scanf("%d %d", &u, &v); G -> AddEdges(u, v); }HLD :: Init(); for(int i = 1, a, b, c, d; i <= q; ++i){scanf("%d %d %d %d", &a, &b, &c, &d); puts(HLD :: Query(a, b, c, d) ? "Y" : "N"); }return 0;
}

谢谢阅读。