CF916E Jamie and Tree 题解

标签：curr int 题解 Jamie return CF916E lca const root

题目链接：CF 或者洛谷

本题难点在于换根 LCA 与换根以后的子树范围寻找，重点讲解

先说操作一，假如原根为 \(1\) 变为了 \(x\)，又变为了 \(y\)，那么其实 \(y\) 和 \(x\) 都可以看做由 \(1\) 变化而来的，即 \(1 \rightarrow x\) 与 \(1 \rightarrow y\)，原因很简单，我们可以把 \(1 \rightarrow x\) 恢复成 \(1\)，再变为 \(y\)，这样换根的形态是没有发生任何变化的。所以这个操作我们可以直接换根。

第二个和第三个操作都可以总结为两步，找以当前为根的 \(LCA\) 与子树在 \(1\) 为根中的实际范围。

先说第一个如何找 \(LCA\)，其实分讨下容易发现，分 \(x\) 与 \(y\) 与 \(root\) 原来是否具有子树关系：

都是 \(root\) 为根的子树上的点，显然 \(lca\) 即为 \(root\)。
一个是子树上的，一个非子树上的。如图所示还是 \(root\)。

都不在子树里。

这个我们这样考虑，\(t_1=lca(root,x)\)，\(t_2=lca(root,y)\)，深度更深的那个，其实就是 lca，原因，这种情况下换根，\(x\) 和 \(y\) 的形态并未发生变化，而 \(t1\) 和 \(t2\) 其实就为以 \(root\) 为根以后的 新子树节点，这个新子树节点包括了 \(x\) 或者 \(y\)。根据第二种换根我们可以知道 \(t1\) 或者 \(t2\) 都有可以是 \(lca\) 的祖先节点，而深度最深的那个显然为真正的 \(lca\)。

这三个情况可以统一成，我们再求出 \(t_3=lca(x,y)\)，那么 \(t_1\)、\(t_2\)、\(t_3\) 中最深的点即为换根后的 \(lca\)，这也是动态 \(lca\) 的基本套路。

接下来解决如何找到当前 \(lca\) 子树范围在原序上的范围。

前两种情况显然 \(lca=root\)，换根后的子树范围即为 \([1,n]\) 包括了整棵树。

考虑第三种情况，分讨下，\(lca\) 和 \(root\) 的关系，显然就两种，\(root\) 是否在 \(lca\) 为根的子树内，一个在它子树当中，一个不在它子树当中。

不在子 \(lca\) 为根的子树内，直接修改原子树即可。

子树形态是并未发生变化的。

在 \(lca\) 为根的子树内，这个比较复杂，如图所示。

不得不说，太类似换根 dp 的套路，这玩意我们容斥来做，整棵树去掉 \(lca\) 下面那个关键点为根的子树贡献即可。

如图所示，全局加然后去掉红色部分贡献，即为正确的了，至于怎么找 \(root \rightarrow lca\) 这条路径上的最后一个点，直接倍增找就行了。

解法一

涉及到了子树加，子树求和，我们使用 \(dfs序+线段树\) 即可，当然区间加区间求和，也可以用差分树状数组，维护两个数组的 bit 即可，这里就用线段树了。当然由于是区间做加法，我们也可以用标记永久化线段树，不过这里就用普通线段树加懒标记即可。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

// #define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e5 + 10;
constexpr int T = 20;
int s[N], e[N], dfn[N], cnt;
int deep[N], fa[N][T + 1];
int a[N], root = 1;
vector<int> child[N];
int n, q;

struct
{
    struct Node
    {
        ll sum, add, len;
    } node[N << 2];

#define len(x) node[x].len
#define add(x) node[x].add
#define sum(x) node[x].sum

    void Add(const int curr, const ll val)
    {
        add(curr) += val;
        sum(curr) += len(curr) * val;
    }

    void pushDown(const int curr)
    {
        if (add(curr))
        {
            Add(ls(curr),add(curr)), Add(rs(curr),add(curr));
            add(curr) = 0;
        }
    }

    void pushUp(const int curr)
    {
        sum(curr) = sum(ls(curr)) + sum(rs(curr));
    }

    void build(const int curr, const int l = 1, const int r = n)
    {
        len(curr) = r - l + 1;
        const int mid = l + r >> 1;
        if (l == r)
        {
            sum(curr) = a[dfn[l]];
            return;
        }
        build(ls(curr), l, mid);
        build(rs(curr), mid + 1, r);
        pushUp(curr);
    }

    void Add(const int curr, const int l, const int r, const int val, const int s = 1, const int e = n)
    {
        if (l <= s and e <= r)
        {
            Add(curr, val);
            return;
        }
        const int mid = s + e >> 1;
        pushDown(curr);
        if (l <= mid)Add(ls(curr), l, r, val, s, mid);
        if (r > mid)Add(rs(curr), l, r, val, mid + 1, e);
        pushUp(curr);
    }

    ll Query(const int curr, const int l, const int r, const int s = 1, const int e = n)
    {
        if (l <= s and e <= r)return sum(curr);
        pushDown(curr);
        const int mid = s + e >> 1;
        ll ans = 0;
        if (l <= mid)ans += Query(ls(curr), l, r, s, mid);
        if (r > mid)ans += Query(rs(curr), l, r, mid + 1, e);
        return ans;
    }
} seg;

inline void dfs(const int curr, const int parent)
{
    deep[curr] = deep[fa[curr][0] = parent] + 1;
    forn(i, 1, T)fa[curr][i] = fa[fa[curr][i - 1]][i - 1];
    dfn[++cnt] = curr;
    s[curr] = cnt;
    for (const auto nxt : child[curr])if (nxt != parent)dfs(nxt, curr);
    e[curr] = cnt;
}

inline int lca(int x, int y)
{
    if (deep[x] < deep[y])swap(x, y);
    forv(i, T, 0)if (deep[fa[x][i]] >= deep[y])x = fa[x][i];
    if (x == y)return x;
    forv(i, T, 0)if (fa[x][i] != fa[y][i])x = fa[x][i], y = fa[y][i];
    return fa[x][0];
}

inline int LCA(const int x, const int y)
{
    const int t1 = lca(root, x), t2 = lca(root, y), t3 = lca(x, y);
    const int maxDeep = max({deep[t1], deep[t2], deep[t3]});
    if (maxDeep == deep[t1])return t1;
    if (maxDeep == deep[t2])return t2;
    return t3;
}

inline int top(int x, int k)
{
    while (k)
    {
        const int step = log2(k);
        x = fa[x][step];
        k -= 1 << step;
    }
    return x;
}

inline void Add(const int curr, const int val)
{
    if (curr == root)seg.Add(1, 1, n, val);
    else if (s[curr] <= s[root] and e[root] <= e[curr])
    {
        const int del = top(root, deep[root] - deep[curr] - 1);
        seg.Add(1, 1, n, val), seg.Add(1, s[del], e[del], -val);
    }
    else seg.Add(1, s[curr], e[curr], val);
}

inline ll Query(const int curr)
{
    if (curr == root)return seg.Query(1, 1, n);
    if (s[curr] <= s[root] and e[root] <= e[curr])
    {
        const int del = top(root, deep[root] - deep[curr] - 1);
        return seg.Query(1, 1, n) - seg.Query(1, s[del], e[del]);
    }
    return seg.Query(1, s[curr], e[curr]);
}

inline void solve()
{
    cin >> n >> q;
    forn(i, 1, n)cin >> a[i];
    forn(i, 1, n-1)
    {
        int u, v;
        cin >> u >> v;
        child[u].push_back(v), child[v].push_back(u);
    }
    dfs(1, 0);
    seg.build(1);
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)cin >> root;
        else if (op == 2)
        {
            int u, v, val;
            cin >> u >> v >> val;
            Add(LCA(u, v), val);
        }
        else
        {
            int curr;
            cin >> curr;
            cout << Query(curr) << endl;
        }
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[时间复杂度为: O((n+q)\log{n}) \]

解法二

参照朋友的博客的换根树剖。其实主要是讲讲 \(top\) 咋求，就是一条路径上的倒数第二个点。

如果 \(root\) 和 \(lca\) 已经在同一条重链上了，显然直接返回 \(lca\) 的重儿子 \(son\)。
不在同一条链让 \(root\) 跳到 \(lca\) 所在重链下面的一条重链的 \(top\) 上。
基于第二天，答案要么为 \(lca\) 的重儿子，要么即为下面一条重链的 \(top\)，主要看 \(fa[root]=lca\)，相同显然即为 \(root\) (跳的结果)，否则为 \(son\)。

参照代码

#include <bits/stdc++.h>

// #pragma GCC optimize(2)
// #pragma GCC optimize("Ofast,no-stack-protector,unroll-loops,fast-math")
// #pragma GCC target("sse,sse2,sse3,ssse3,sse4.1,sse4.2,avx,avx2,popcnt,tune=native")

// #define isPbdsFile

#ifdef isPbdsFile

#include <bits/extc++.h>

#else

#include <ext/pb_ds/priority_queue.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/tree_policy.hpp>
#include <ext/pb_ds/trie_policy.hpp>
#include <ext/pb_ds/tag_and_trait.hpp>
#include <ext/pb_ds/hash_policy.hpp>
#include <ext/pb_ds/list_update_policy.hpp>
#include <ext/pb_ds/assoc_container.hpp>
#include <ext/pb_ds/exception.hpp>
#include <ext/rope>

#endif

using namespace std;
using namespace __gnu_cxx;
using namespace __gnu_pbds;
typedef long long ll;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef tuple<int, int, int> tii;
typedef tuple<ll, ll, ll> tll;
typedef unsigned int ui;
typedef unsigned long long ull;
typedef __int128 i128;
#define hash1 unordered_map
#define hash2 gp_hash_table
#define hash3 cc_hash_table
#define stdHeap std::priority_queue
#define pbdsHeap __gnu_pbds::priority_queue
#define sortArr(a, n) sort(a+1,a+n+1)
#define all(v) v.begin(),v.end()
#define yes cout<<"YES"
#define no cout<<"NO"
#define Spider ios_base::sync_with_stdio(false);cin.tie(nullptr);cout.tie(nullptr);
#define MyFile freopen("..\\input.txt", "r", stdin),freopen("..\\output.txt", "w", stdout);
#define forn(i, a, b) for(int i = a; i <= b; i++)
#define forv(i, a, b) for(int i=a;i>=b;i--)
#define ls(x) (x<<1)
#define rs(x) (x<<1|1)
#define endl '\n'
//用于Miller-Rabin
[[maybe_unused]] static int Prime_Number[13] = {0, 2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, 37};

template <typename T>
int disc(T* a, int n)
{
    return unique(a + 1, a + n + 1) - (a + 1);
}

template <typename T>
T lowBit(T x)
{
    return x & -x;
}

template <typename T>
T Rand(T l, T r)
{
    static mt19937 Rand(time(nullptr));
    uniform_int_distribution<T> dis(l, r);
    return dis(Rand);
}

template <typename T1, typename T2>
T1 modt(T1 a, T2 b)
{
    return (a % b + b) % b;
}

template <typename T1, typename T2, typename T3>
T1 qPow(T1 a, T2 b, T3 c)
{
    a %= c;
    T1 ans = 1;
    for (; b; b >>= 1, (a *= a) %= c)if (b & 1)(ans *= a) %= c;
    return modt(ans, c);
}

template <typename T>
void read(T& x)
{
    x = 0;
    T sign = 1;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')sign = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        x = (x << 3) + (x << 1) + (ch ^ 48);
        ch = getchar();
    }
    x *= sign;
}

template <typename T, typename... U>
void read(T& x, U&... y)
{
    read(x);
    read(y...);
}

template <typename T>
void write(T x)
{
    if (typeid(x) == typeid(char))return;
    if (x < 0)x = -x, putchar('-');
    if (x > 9)write(x / 10);
    putchar(x % 10 ^ 48);
}

template <typename C, typename T, typename... U>
void write(C c, T x, U... y)
{
    write(x), putchar(c);
    write(c, y...);
}


template <typename T11, typename T22, typename T33>
struct T3
{
    T11 one;
    T22 tow;
    T33 three;

    bool operator<(const T3 other) const
    {
        if (one == other.one)
        {
            if (tow == other.tow)return three < other.three;
            return tow < other.tow;
        }
        return one < other.one;
    }

    T3() { one = tow = three = 0; }

    T3(T11 one, T22 tow, T33 three) : one(one), tow(tow), three(three)
    {
    }
};

template <typename T1, typename T2>
void uMax(T1& x, T2 y)
{
    if (x < y)x = y;
}

template <typename T1, typename T2>
void uMin(T1& x, T2 y)
{
    if (x > y)x = y;
}

constexpr int N = 1e5 + 10;
constexpr int T = 20;
int deep[N], son[N], siz[N], fa[N];
vector<int> child[N];
int n, q;

inline void dfs1(const int curr, const int parent)
{
    deep[curr] = deep[fa[curr] = parent] + 1;
    siz[curr] = 1;
    for (const auto nxt : child[curr])
    {
        if (nxt == parent)continue;
        dfs1(nxt, curr);
        siz[curr] += siz[nxt];
        if (siz[nxt] > siz[son[curr]])son[curr] = nxt;
    }
}

int idx[N], s[N], e[N], top[N], cnt;

inline void dfs2(const int curr, const int root)
{
    idx[++cnt] = curr, s[curr] = cnt, e[curr] = s[curr] + siz[curr] - 1;
    top[curr] = root;
    if (son[curr])dfs2(son[curr], root);
    for (const auto nxt : child[curr])if (nxt != fa[curr] and nxt != son[curr])dfs2(nxt, nxt);
}

int a[N], root = 1;

struct
{
    struct Node
    {
        ll sum, add, len;
    } node[N << 2];

#define len(x) node[x].len
#define add(x) node[x].add
#define sum(x) node[x].sum

    void Add(const int curr, const ll val)
    {
        add(curr) += val;
        sum(curr) += len(curr) * val;
    }

    void pushDown(const int curr)
    {
        if (add(curr))
        {
            Add(ls(curr),add(curr)), Add(rs(curr),add(curr));
            add(curr) = 0;
        }
    }

    void pushUp(const int curr)
    {
        sum(curr) = sum(ls(curr)) + sum(rs(curr));
    }

    void build(const int curr, const int l = 1, const int r = n)
    {
        len(curr) = r - l + 1;
        const int mid = l + r >> 1;
        if (l == r)
        {
            sum(curr) = a[idx[l]];
            return;
        }
        build(ls(curr), l, mid);
        build(rs(curr), mid + 1, r);
        pushUp(curr);
    }

    void Add(const int curr, const int l, const int r, const int val, const int s = 1, const int e = n)
    {
        if (l <= s and e <= r)
        {
            Add(curr, val);
            return;
        }
        const int mid = s + e >> 1;
        pushDown(curr);
        if (l <= mid)Add(ls(curr), l, r, val, s, mid);
        if (r > mid)Add(rs(curr), l, r, val, mid + 1, e);
        pushUp(curr);
    }

    ll Query(const int curr, const int l, const int r, const int s = 1, const int e = n)
    {
        if (l <= s and e <= r)return sum(curr);
        pushDown(curr);
        const int mid = s + e >> 1;
        ll ans = 0;
        if (l <= mid)ans += Query(ls(curr), l, r, s, mid);
        if (r > mid)ans += Query(rs(curr), l, r, mid + 1, e);
        return ans;
    }
} seg;

inline int lca(int x, int y)
{
    while (top[x] != top[y])
    {
        if (deep[top[x]] < deep[top[y]])swap(x, y);
        x = fa[top[x]];
    }
    if (deep[x] > deep[y])swap(x, y);
    return x;
}

inline int LCA(const int x, const int y)
{
    const int t1 = lca(root, x), t2 = lca(root, y), t3 = lca(x, y);
    const int maxDeep = max({deep[t1], deep[t2], deep[t3]});
    if (maxDeep == deep[t1])return t1;
    if (maxDeep == deep[t2])return t2;
    return t3;
}

//从root到curr的路径上最后一个点
inline int TreeTop(const int curr, int x = root)
{
    if (top[curr] == top[x])return son[curr];
    while (top[fa[top[x]]] != top[curr])x = fa[top[x]];
    x = top[x];
    if (fa[x] != curr)x = son[curr];
    return x;
}

inline bool sameTree(const int curr)
{
    return s[curr] <= s[root] and e[root] <= e[curr];
}

inline void Add(const int curr, const int val)
{
    if (curr == root)seg.Add(1, 1, n, val);
    else if (sameTree(curr))
    {
        const int del = TreeTop(curr);
        seg.Add(1, 1, n, val), seg.Add(1, s[del], e[del], -val);
    }
    else seg.Add(1, s[curr], e[curr], val);
}

inline ll Query(const int curr)
{
    if (curr == root)return seg.Query(1, 1, n);
    if (sameTree(curr))
    {
        const int del = TreeTop(curr);
        return seg.Query(1, 1, n) - seg.Query(1, s[del], e[del]);
    }
    return seg.Query(1, s[curr], e[curr]);
}

inline void solve()
{
    cin >> n >> q;
    forn(i, 1, n)cin >> a[i];
    forn(i, 1, n-1)
    {
        int u, v;
        cin >> u >> v;
        child[u].push_back(v), child[v].push_back(u);
    }
    dfs1(1, 0);
    dfs2(1, 1);
    seg.build(1);
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)cin >> root;
        else if (op == 2)
        {
            int u, v, val;
            cin >> u >> v >> val;
            Add(LCA(u, v), val);
        }
        else
        {
            int curr;
            cin >> curr;
            cout << Query(curr) << endl;
        }
    }
}

signed int main()
{
    // MyFile
    Spider
    //------------------------------------------------------
    // clock_t start = clock();
    int test = 1;
    //    read(test);
    // cin >> test;
    forn(i, 1, test)solve();
    //    while (cin >> n, n)solve();
    //    while (cin >> test)solve();
    // clock_t end = clock();
    // cerr << "time = " << double(end - start) / CLOCKS_PER_SEC << "s" << endl;
}

\[不涉及跳链修改与查询，只有子树操作，时间复杂度为:\ O((n+q)\log{n}) \]

如果要用 \(LCT\) 做子树操作是较为困难的，当然也能做，具体的每个点同时维护虚子树信息，但这些信息不能单纯地使用懒标记维护，需要维护一个标记永久化的虚子树全局加标记，在虚实变化时通过这个标记来更新真实的虚实子树和总信息，细节较多，后续写了再补代码。当然 \(ETT\) 和 \(Top Tree\) 应该也是完全能做的。

标签：curr,int,题解,Jamie,return,CF916E,lca,const,root
From： https://www.cnblogs.com/Athanasy/p/18029516

题目链接：CF 或者洛谷

本题难点在于换根 LCA 与换根以后的子树范围寻找，重点讲解

解法一

解法二

相关文章

赞助商

阅读排行

CF916E Jamie and Tree 题解

题目链接：CF 或者 洛谷

本题难点在于换根 LCA 与换根以后的子树范围寻找，重点讲解

解法一

解法二

相关文章

赞助商

阅读排行

题目链接：CF 或者洛谷