Codeforces 852I Dating 树上莫队

Dating

随便树上莫队搞一搞就好啦。

#include<bits/stdc++.h>
#define LL long long
#define LD long double
#define ull unsigned long long
#define fi first
#define se second
#define mk make_pair
#define PLL pair<LL, LL>
#define PLI pair<LL, int>
#define PII pair<int, int>
#define SZ(x) ((int)x.size())
#define ALL(x) (x).begin(), (x).end()
#define fio ios::sync_with_stdio(false); cin.tie(0);

using namespace std;

const int N = 2e5 + 7;
const int inf = 0x3f3f3f3f;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const int mod = 1e9 + 7;
const double eps = 1e-8;
const double PI = acos(-1);

template<class T, class S> inline void add(T& a, S b) {a += b; if(a >= mod) a -= mod;}
template<class T, class S> inline void sub(T& a, S b) {a -= b; if(a < 0) a += mod;}
template<class T, class S> inline bool chkmax(T& a, S b) {return a < b ? a = b, true : false;}
template<class T, class S> inline bool chkmin(T& a, S b) {return a > b ? a = b, true : false;}


const int B = 500;

int n, q, depth[N], f[N], pa[N][20], gender[N];
int in[N], out[N], id[N], idx, op[N];
LL ans[N];
bool flag[N];
int cnt[2][N];

vector<int> oo;
vector<int> G[N];

struct Qus {
    int L, R, lca, id;
    bool operator < (const Qus& rhs) const {
        if(L / B == rhs.L / B) return R < rhs.R;
        return L < rhs.L;
    }
} qus[N];


int l, r; LL ret;

inline void update(int x) {
    ret += op[id[x]] * cnt[gender[id[x]] ^ 1][f[id[x]]];
    cnt[gender[id[x]]][f[id[x]]] += op[id[x]];
    op[id[x]] = -op[id[x]];
}

void dfs(int u, int fa) {
    depth[u] = depth[fa] + 1;
    pa[u][0] = fa;
    for(int i = 1; i < 20; i++)
        pa[u][i] = pa[pa[u][i - 1]][i - 1];
    id[++idx] = u;
    in[u] = idx;
    for(auto& v : G[u]) {
        if(v == fa) continue;
        dfs(v, u);
    }
    id[++idx] = u;
    out[u] = idx;
}

int getLca(int u, int v) {
    if(depth[u] < depth[v]) swap(u, v);
    for(int i = 19; ~i; i--)
        if((depth[u] - depth[v]) >> i & 1)
            u = pa[u][i];
    if(u == v) return u;
    for(int i = 19; ~i; i--)
        if(pa[u][i] != pa[v][i])
            u = pa[u][i], v = pa[v][i];
    return pa[u][0];
}

int main() {
    scanf("%d", &n);
    for(int i = 1; i <= n; i++) scanf("%d", &gender[i]), op[i] = 1;
    for(int i = 1; i <= n; i++) {
        scanf("%d", &f[i]);
        oo.push_back(f[i]);
    }
    for(int i = 1; i < n; i++) {
        int a, b; scanf("%d%d", &a, &b);
        G[a].push_back(b);
        G[b].push_back(a);
    }
    sort(ALL(oo));
    oo.erase(unique(ALL(oo)), oo.end());
    for(int i = 1; i <= n; i++)
        f[i] = lower_bound(ALL(oo), f[i]) - oo.begin();
    dfs(1, 0);
    scanf("%d", &q);
    for(int i = 1; i <= q; i++) {
        int a, b, lca;
        scanf("%d%d", &a, &b);
        lca = getLca(a, b);
        if(lca == a || lca == b) {
            if(a == lca) qus[i] = Qus{in[a], in[b], a, i};
            else qus[i] = Qus{in[b], in[a], b, i};
        } else {
            if(in[a] < in[b]) qus[i] = Qus{out[a], in[b], lca, i};
            else qus[i] = Qus{out[b], in[a], lca, i};
        }
    }
    l = 1, r = 0, ret = 0;
    sort(qus + 1, qus + 1 + q);
    for(int o = 1; o <= q; o++) {
        int L = qus[o].L, R = qus[o].R, lca = qus[o].lca, who = qus[o].id;
        while(r < R) update(++r);
        while(l > L) update(--l);
        while(r > R) update(r--);
        while(l < L) update(l++);
        if(lca != id[L] && lca != id[R]) ans[who] = ret + cnt[gender[lca] ^ 1][f[lca]];
        else ans[who] = ret;
    }
    for(int i = 1; i <= q; i++) printf("%lld\n", ans[i]);
    return 0;
}

/*
*/