CodeForces 519E 树形DP A and B and Lecture Rooms

给出一棵树，有若干次询问，每次询问距两个点u, v距离相等的点的个数。

情况还挺多的，少侠不妨去看官方题解。^_^

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <vector>
using namespace std;

const int maxn = 100000 + 10;

int n, Q;
vector<int> G[maxn];

int L[maxn];
int fa[maxn];
int sz[maxn];

void dfs(int u)
{
    sz[u] = 1;
    for(int i = 0; i < G[u].size(); i++)
    {
        int v = G[u][i];
        if(v == fa[u]) continue;
        fa[v] = u;
        L[v] = L[u] + 1;
        dfs(v);
        sz[u] += sz[v];
    }
}

const int logmaxn = 20;
int anc[maxn][logmaxn];

int ancestor(int a, int x)
{
    if(!x) return a;
    for(int i = 0; (1 << i) <= x; i++)
        if(x & (1 << i)) a = anc[a][i];

    return a;
}

int LCA(int p, int q)
{
    if(L[p] < L[q]) swap(p, q);
    int log;
    for(log = 1; (1 << log) <= L[p]; log++); log--;
    for(int i = log; i >= 0; i--)
        if(L[p] - (1 << i) >= L[q]) p = anc[p][i];
    if(p == q) return p;
    for(int i = log; i >= 0; i--)
        if(anc[p][i] && anc[p][i] != anc[q][i])
            p = anc[p][i], q = anc[q][i];
    return fa[p];
}

int main()
{
    //freopen("in.txt", "r", stdin);

    scanf("%d", &n);
    for(int i = 1; i < n; i++)
    {
        int u, v; scanf("%d%d", &u, &v);
        G[u].push_back(v);
        G[v].push_back(u);
    }

    dfs(1);

    for(int i = 1; i <= n; i++) anc[i][0] = fa[i];
    for(int j = 1; (1 << j) < n; j++)
        for(int i = 1; i <= n; i++) if(anc[i][j-1])
            anc[i][j] = anc[anc[i][j-1]][j-1];

    scanf("%d", &Q);
    while(Q--)
    {
        int u, v; scanf("%d%d", &u, &v);

        if(u == v) { printf("%d
", n); continue; }

        int l = LCA(u, v);

        if((L[u] + L[v] - L[l] * 2) & 1) { puts("0"); continue; }

        if(L[u] == L[v])
        {
            int t = L[u] - L[l];
            int uu = ancestor(u, t - 1), vv = ancestor(v, t - 1);
            printf("%d
", n - sz[uu] - sz[vv]);
            continue;
        }

        if(L[u] < L[v]) swap(u, v);
        int t = L[u] + L[v] - L[l] * 2;
        int p1 = ancestor(u, t / 2);
        int p2 = ancestor(u, t / 2 - 1);
        printf("%d
", sz[p1] - sz[p2]);
    }

    return 0;
}