CF383C Propagating tree 题解

合集 - codeforces题解集1(37)

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28.CF383C Propagating tree 题解2024-02-29

29.CF1436E Complicated Computations 题解2024-03-08 30.CF817F MEX Queries 题解2024-03-20 31.CF1638E Colorful Operations 题解2024-03-21 32.CF1618G Trader Problem 题解2024-03-22 33.CF794F Leha and security system 题解2024-03-23 34.CF371E Subway Innovation 题解2024-04-04 35.CF1200E Compress Words 题解2024-04-06 36.CF1884D Counting Rhyme 题解2024-05-17 37.CF1982F Sorting Problem Again 题解2024-06-26

题目链接：CF 或者洛谷

比较朴素的题，注意到这个涉及到子树变化，我们考虑优先处理出 $d f s$ 序，方便处理。

注意到第一个问题较为繁琐，我们着重解决下第一个问题。在树上问题，我们这种间隔点常常使用 $d e e p$ 进行区分。

根的 $d e e p$ 为奇数，那么对自己子树范围内的奇数 $d e e p$ 都是 $+ v a l$ ，偶数 $d e e p$ 都是 $- v a l$ 。
根的 $d e e p$ 为偶数，那么对自己子树范围内的偶数 $d e e p$ 都是 $+ v a l$ ，奇数 $d e e p$ 都是 $- v a l$ 。

为此，比较直接的想法，我们直接分两种贡献，分别维护两个 $B I T$ 进行整棵子树变化，但在询问答案时，对于当前点的 $d e e p$ 分讨，以奇数为例，那么第一个 $B I T$ 的贡献即为 $+$ ，第二个 $B I T$ 即为 $-$ 。如果为偶数深度，显然就反过来了。关于子树加和询问单点，我们使用树状数组维护差分答案即可。

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 = 2e5 + 10;
int n, q;
 
struct
{
    int bit[N];
 
    void add(int x, const int val)
    {
        while (x <= n)bit[x] += val, x += lowBit(x);
    }
 
    int query(int x) const
    {
        int ans = 0;
        while (x)ans += bit[x], x -= lowBit(x);
        return ans;
    }
} bitOdd, bitEven;
 
vector<int> child[N];
int s[N], e[N], deep[N], cnt;
 
inline void dfs(const int curr, const int fa)
{
    deep[curr] = deep[fa] + 1;
    s[curr] = ++cnt;
    for (const auto nxt : child[curr])if (nxt != fa)dfs(nxt, curr);
    e[curr] = cnt;
}
 
int a[N];
 
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);
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int u, val;
            cin >> u >> val;
            if (deep[u] & 1)bitOdd.add(s[u], val), bitOdd.add(e[u] + 1, -val);
            else bitEven.add(s[u], val), bitEven.add(e[u] + 1, -val);
        }
        else
        {
            int u;
            cin >> u;
            int odd = bitOdd.query(s[u]), even = bitEven.query(s[u]);
            cout << (deep[u] & 1 ? a[u] + odd - even : a[u] + even - odd) << 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;
}

考虑下优化，我们的两个 $B I T$ 能否合成一个？以奇数深度为例，我们注意到最终对于奇数深度的点，询问的答案为 $a [i] + o d d - e v e n$ ，那么我们 $B I T$ 直接奇数深度的点还是维护 $+$ ，偶数深度直接维护 $-$ ，这样直接查出来的答案就是奇数深度点的答案。偶数深度询问点答案，我们注意到为 $a [i] - e v e n + o d d$ ，即为加上它的相反数，这样一来我们就优化到只用需要一个 $B I T$ 就能统计答案了。

优化参照代码

 #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 = 2e5 + 10;
int n, q;
 
 
int bit[N];
 
void add(int x, const int val)
{
    while (x <= n)bit[x] += val, x += lowBit(x);
}
 
int query(int x)
{
    int ans = 0;
    while (x)ans += bit[x], x -= lowBit(x);
    return ans;
}
 
 
vector<int> child[N];
int s[N], e[N], deep[N], cnt;
 
inline void dfs(const int curr, const int fa)
{
    deep[curr] = deep[fa] + 1;
    s[curr] = ++cnt;
    for (const auto nxt : child[curr])if (nxt != fa)dfs(nxt, curr);
    e[curr] = cnt;
}
 
int a[N];
 
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);
    while (q--)
    {
        int op;
        cin >> op;
        if (op == 1)
        {
            int u, val;
            cin >> u >> val;
            if (!(deep[u] & 1))val = -val;
            add(s[u], val), add(e[u] + 1, -val);
        }
        else
        {
            int u;
            cin >> u;
            int val = query(s[u]);
            if (!(deep[u] & 1))val = -val;
            cout << a[u] + val << 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 (q \log n)

posted @ 2024-02-29 09:48 Athanasy 阅读(48) 评论(0) 编辑收藏举报

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CF383C Propagating tree 题解

题目链接：CF 或者洛谷

比较朴素的题，注意到这个涉及到子树变化，我们考虑优先处理出 $d f s$ 序，方便处理。

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	#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 = 2e5 + 10;
	int n, q;

	struct
	{
	int bit[N];

	void add(int x, const int val)
	{
	while (x <= n)bit[x] += val, x += lowBit(x);
	}

	int query(int x) const
	{
	int ans = 0;
	while (x)ans += bit[x], x -= lowBit(x);
	return ans;
	}
	} bitOdd, bitEven;

	vector<int> child[N];
	int s[N], e[N], deep[N], cnt;

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

	int a[N];

	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);
	while (q--)
	{
	int op;
	cin >> op;
	if (op == 1)
	{
	int u, val;
	cin >> u >> val;
	if (deep[u] & 1)bitOdd.add(s[u], val), bitOdd.add(e[u] + 1, -val);
	else bitEven.add(s[u], val), bitEven.add(e[u] + 1, -val);
	}
	else
	{
	int u;
	cin >> u;
	int odd = bitOdd.query(s[u]), even = bitEven.query(s[u]);
	cout << (deep[u] & 1 ? a[u] + odd - even : a[u] + even - odd) << 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;
	}

Athanasy

CF383C Propagating tree 题解

题目链接：CF 或者 洛谷

比较朴素的题，注意到这个涉及到子树变化，我们考虑优先处理出 dfs 序，方便处理。

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题目链接：CF 或者洛谷

比较朴素的题，注意到这个涉及到子树变化，我们考虑优先处理出 $d f s$ 序，方便处理。