2021牛客暑期多校训练营7 F.xay loves trees 主席树+dfs序

Posted 2021-09-03 kaka0010

原题链接：https://ac.nowcoder.com/acm/contest/11258/F

目录

• 题意
• 分析
• Code

题意

给定两棵树，要求找一个点集使得两棵树上都满足：

1. 第一棵树上点集是相互连通的，而且任意两点互为祖先节点
2. 第二棵树上任意两点都不能互为祖先节点

分析

首先这道题我们必须知道一个性质，一棵树上的dfs序入点和出点一定是连续的，且子树内的dfs序一定大于根节点的dfs序。这样其实就变成了我们在dfs序上找最多的连续区间，使得他们不相交。但要满足第一棵树上连续成链的性质，其实不难想到固定一个端点然后二分去找最远取到的祖先节点。

至于在树上维护一条链，我们在遍历点的时候将点压入栈里，回溯的时候出栈就可以了，这时这个栈内的点一定是一条链上且连续，我们在每个节点上建立一个主席树，每次选择一个右端点，将右端点贡献加入主席树内，至于贡献怎么算？只要将当前点的入点dfs序和出点dfs序的区间+1，这样之后每次查询当前端点子树内是否被染色过来进行二分。

光这样还不够，每次二分的区间也必须有限制，我们要记录每次符合要求的最右端，并将二分的最左端设置为这个值，因为答案具有单调性。

Code

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int ul;
typedef pair<int, int> PII;
const ll inf = 1e18;
const int N = 3e5 + 10;
const int M = 1e6 + 10;
const ll mod = 1e9 + 7;
const double eps = 1e-8;

#define lowbit(i) (i & -i)
#define Debug(x) cout << (x) << endl
#define fi first
#define se second
#define mem memset
#define endl '\\n'

namespace StandardIO {
    template<typename T>
    inline void read(T &x) {
        x = 0; T f = 1;
        char c = getchar();
        for (; c < '0' || c > '9'; c = getchar()) if (c == '-') f = -1;
        for (; c >= '0' && c <= '9'; c = getchar()) x = x * 10 + c - '0';
        x *= f;
    }

    template<typename T>
    inline void write(T x) {
        if (x < 0) putchar('-'), x *= -1;
        if (x >= 10) write(x / 10);
        putchar(x % 10 + '0');
    }
}
namespace Comb {
    ll f[N];
    ll ksm(ll a, ll b) {
        ll res = 1, base = a;
        while (b) {
            if (b & 1) res = res * base % mod;
            base = base * base % mod;
            b >>= 1;
        }
        return res;
    }
    void init() {
        f[0] = 1;
        for (ll i = 1; i < N; i++) f[i] = f[i - 1] * i % mod;
    }
    ll C(ll a, ll b) {
        if (a < 0 || b < 0 || b > a) return 0;
        return f[a] * ksm(f[a - b], mod - 2) % mod * ksm(f[b], mod - 2) % mod;
    }
}

struct node {
    int ls, rs;
    int sum, tag;
}hjt[N*50];
int cnt, rt[N];
void modify(int &now, int pre, int ql, int qr, int l, int r, int val) {
    now = ++cnt;
    hjt[now] = hjt[pre];
    hjt[now].sum += (min(qr, r) - max(ql, l) + 1) * val;
    if (ql <= l && qr >= r) {
        hjt[now].tag += val;
        return;
    }
    int mid = (l + r) >> 1;
    if (ql <= mid) modify(hjt[now].ls, hjt[pre].ls, ql, qr, l, mid, val);
    if (qr > mid) modify(hjt[now].rs, hjt[pre].rs, ql, qr, mid+1, r, val);
}
int query(int now, int pre, int ql, int qr, int l, int r) {
    if (ql <= l && qr >= r) return hjt[now].sum - hjt[pre].sum;
    int mid = (l + r) >> 1;
    int ans = (min(qr, r) - max(ql, l) + 1) * (hjt[now].tag - hjt[pre].tag);
    if (ql <= mid) ans += query(hjt[now].ls, hjt[pre].ls, ql, qr, l, mid);
    if (qr > mid) ans += query(hjt[now].rs, hjt[pre].rs, ql, qr, mid+1, r);
    return ans;
}
vector<int> G1[N], G2[N];
int n, in[N], out[N], tot;
void dfs1(int x, int fa) {
    in[x] = ++tot;
    for (auto v : G2[x]) {
        if (v == fa) continue;
        dfs1(v, x);
    }
    out[x] = tot;
}
int stk[N], top, ans, pre[N];
void dfs2(int x, int fa) {
    stk[++top] = x; rt[x] = rt[fa];
    int l = pre[fa], r = top;
    while (l <= r) {
        int mid = (l + r) >> 1;
        //cout << x << ' ' << stk[mid] << ' ' << query(rt[x], rt[stk[mid]], in[x], out[x], 1, n) << endl;
        if (query(rt[x], rt[stk[mid]], in[x], out[x], 1, n)) l = mid + 1;
        else r = mid - 1;
    }
    //cout << x << ' ' << top << ' ' << top - l << ' ' << pre[fa] << endl;
    pre[x] = l;
    ans = max(ans, top - l);
    modify(rt[x], rt[x], in[x], out[x], 1, n, 1);
    for (auto v : G1[x]) {
        if (v == fa) continue;
        dfs2(v, x);
    }
    top--;
}
void init() {
    for (int i = 1; i <= n; i++) {
        G1[i].clear();
        G2[i].clear();
        in[i] = out[i] = 0;
        rt[i] = 0;
    }
    for (int i = 1; i <= cnt; i++) hjt[i].ls = hjt[i].rs = hjt[i].sum = hjt[i].tag = 0;
    cnt = tot = top = 0;
    ans = 1;
}
inline void solve() {
    int T; cin >> T; while (T--) {
        cin >> n;
        init();
        for (int i = 1; i <= n - 1; i++) {
            int u, v;
            cin >> u >> v;
            G1[u].push_back(v);
            G1[v].push_back(u);
        }
        for (int i = 1; i <= n - 1; i++) {
            int u, v;
            cin >> u >> v;
            G2[u].push_back(v);
            G2[v].push_back(u);
        }
        dfs1(1, 0);
        dfs2(1, 0);
        cout << ans << endl;
    }
}

signed main() {
    ios_base::sync_with_stdio(false);
    cin.tie(0);
    cout.tie(0);
#ifdef ACM_LOCAL
    freopen("input", "r", stdin);
    freopen("output", "w", stdout);
    signed test_index_for_debug = 1;
    char acm_local_for_debug = 0;
    do {
        if (acm_local_for_debug == '$') exit(0);
        if (test_index_for_debug > 20)
            throw runtime_error("Check the stdin!!!");
        auto start_clock_for_debug = clock();
        solve();
        auto end_clock_for_debug = clock();
        cout << "Test " << test_index_for_debug << " successful" << endl;
        cerr << "Test " << test_index_for_debug++ << " Run Time: "
             << double(end_clock_for_debug - start_clock_for_debug) / CLOCKS_PER_SEC << "s" << endl;
        cout << "--------------------------------------------------" << endl;
    } while (cin >> acm_local_for_debug && cin.putback(acm_local_for_debug));
#else
    solve();
#endif
    return 0;
}