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#include <bits/stdc++.h>
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using namespace std;
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const int N = 40010, M = 80010;
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//链式前向星
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int e[M], h[N], idx, ne[M];
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void add(int a, int b) {
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e[idx] = b, ne[idx] = h[a], h[a] = idx++;
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}
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int n;
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int depth[N];
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int f[N][16];
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//树上深搜
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void dfs(int u) {
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for (int i = h[u]; ~i; i = ne[i]) {
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int j = e[i];
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if (!depth[j]) {
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depth[j] = depth[u] + 1; //① 每个节点的深度
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f[j][0] = u; //② j的父节点是u
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dfs(j);
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}
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}
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}
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int lca(int a, int b) {
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if (depth[a] < depth[b]) swap(a, b);
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for (int k = 15; k >= 0; k--)
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if (depth[f[a][k]] >= depth[b]) a = f[a][k];
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if (a == b) return a;
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return f[a][0];
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}
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int main() {
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memset(h, -1, sizeof h); //初始化链式前向星
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scanf("%d", &n); // n个顶点
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int a, b, m, root;
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while (n--) {
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scanf("%d %d", &a, &b); //表示a和b之间有一条无向边。如果b是−1,那么a就是树的根
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if (b == -1)
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root = a;
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else
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add(a, b), add(b, a); //有根无向树
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}
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//从根开始进行处理,根的深度是1
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depth[root] = 1;
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dfs(root); //从根开始深搜,目标是 ①:给每个节点标识上深度 ②:给每个节点设置上打表f[i][0]的值,其实就是他爹是谁
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// dfs只解决了2^0问题,其它的2^1,2^2,2^3,...,2^k,还需要再写循环,自己实现,这个顺序与bfs不一样,bfs根据深度枚举,遇到i时,它的依赖前序都已准备好, dfs需要全跑完再来一遍
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// 这个枚举序挺烦人啊,dp枚举顺序是配合业务来的,从这个来看,固定好2^i,然后逐个枚举每个点号进行填充
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// 所以,yxc大佬没有讲按dfs进行标识深度,而是采用了bfs进行标识,bfs是很好理解
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for (int i = 1; i <= 15; i++)
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for (int j = 1; j <= N; j++)
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f[j][i] = f[f[j][i - 1]][i - 1];
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scanf("%d", &m);
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while (m--) {
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scanf("%d %d", &a, &b);
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int p = lca(a, b);
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if (p == a)
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puts("1");
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else if (p == b)
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puts("2");
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else
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puts("0");
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}
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return 0;
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}
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