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Code in c++

Creado el: 13 de julio de 2025

Respondido usando O3 por Chat01

Pregunta

17524086886888569450647906673485.jpg17524086886888569450647906673485.jpg

Code in c++

Pensamiento

Interpreting C++ task

The user provided an image with a game problem involving Alice and Bob on a tree. Alice chooses a node and sets its value to 0, and Bob moves to a child node, aiming to maximize the score. The score depends on the node Bob stops at. The game ends when Bob reaches a leaf or decides to stop. I think I’ll need to write a C++ solution that computes the score recursively, considering optimal plays from both Alice and Bob at each node. I’m excited to dig into the algorithm!

Clarifying the game setup

Bob has the choice to stop at his current node, which gives him control over when to maximize his score. The game involves Alice setting a node's value to 0, and Bob then moving to one of the child nodes, always aiming to maximize his score. The game ends when he reaches a leaf or decides to stop. The goal is to compute the minimax value. To handle the input, I'll parse the tree size, edges list, and node values, then apply the algorithm based on alternating moves.

Breaking down the game mechanics

In this game, Alice can set a node's value to 0 multiple times, and Bob moves along the tree trying to maximize his score, which is impacted by Alice's actions. The game continues until Bob decides to stop or reaches a leaf. Each turn, Alice zeros a node, and Bob chooses a path accordingly. Bob’s goal is to maximize his score by choosing children or stopping, while Alice aims to minimize it. Solving this optimally seems to involve dynamic programming, considering both players’ strategies at each node and depth.

Simplifying the problem

The game seems to work by Alice zeroing a node at the start, then Bob pathing through the tree to maximize his score. Alice's goal is to minimize Bob's score by zeroing the highest value node. Bob can only choose a downward path through the tree and stops when he reaches a leaf or decides to stop earlier. The main strategy is for Alice to zero the highest value node, which makes Bob aim for the second-highest value node, considering any potential duplicates. Edge cases include when the tree has only one node.

Thought for 44 seconds

Respuesta

Below is a straightforward C++ solution that follows the minimax logic described in the photographed statement.
Because Alice may set exactly one node’s value to 0 before Bob ever moves, her optimal play is to wipe out the largest value she can find.
Bob then simply walks down the tree (or stops immediately) and chooses the node with the greatest remaining value on any root-to-leaf path.
Thus:

  • If the largest value occurs more than once, Alice can only erase one copy, so Bob can still reach a node with that same largest value — the score remains that value.
  • If the largest value occurs exactly once, Alice erases it, so Bob’s best achievable value is the second-largest value (or 0 if the tree held only one vertex).

That reduces the game to a simple “largest / second-largest” computation that is O(N) time and O(1) extra space.

cpp
#include <bits/stdc++.h>
using namespace std;

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    long long N, M, Two;
    if (!(cin >> N)) return 0;          // no input
    cin >> M >> Two;                    // Two should be 2, but we don’t need it

    vector<long long> A(N);
    for (long long &x : A) cin >> x;

    // Read and discard the edge list; it is irrelevant to the final score
    for (long long i = 0, u, v; i < M; ++i) cin >> u >> v;

    // Find max1 (largest), its frequency, and max2 (second-largest)
    long long max1 = LLONG_MIN, max2 = LLONG_MIN;
    long long cntMax1 = 0;

    for (long long val : A) {
        if (val > max1) {
            max2 = max1;
            max1 = val;
            cntMax1 = 1;
        } else if (val == max1) {
            ++cntMax1;
        } else if (val > max2) {
            max2 = val;
        }
    }

    long long answer;
    if (cntMax1 >= 2)          // Alice can’t wipe out every copy of the max
        answer = max1;
    else                       // unique maximum gets zeroed
        answer = (max2 == LLONG_MIN ? 0 : max2);

    cout << answer << '\n';
    return 0;
}

Why this is correct

  1. Alice’s best move is always to eliminate the current global maximum; any other choice can’t yield a lower eventual score.
  2. Bob’s best move is then to take the maximum value still present along a path from the root (and every vertex is on some such path), so the score equals the greatest remaining value.
  3. The two bullet points above reduce the game to the simple “largest vs. second-largest” comparison implemented.

This meets the problem’s requirements in O(N) time and uses O(1) auxiliary memory.

Pregunta

Case 3

Input:

32

2

2

10

14

19

12

23

Output:

10

Explanation:

N-3, M-2, Two-2, A-[10,14,19], Edges-[[1,2], [2,3]] Bob will end the game in node 1.

Python

python
from PIL import Image
import matplotlib.pyplot as plt

img = Image.open('/mnt/data/17524086886888569450647906673485.jpg')
plt.imshow(img); plt.axis('off')

(-0.5, 4607.5, 3455.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img.crop((200,120,1000,1500))); plt.axis('off')

(-0.5, 799.5, 1379.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img.crop((300,0,2000,1500))); plt.axis('off')

(-0.5, 1699.5, 1499.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img.crop((300,400,2000,2000))); plt.axis('off')

(-0.5, 1699.5, 1599.5, -0.5)

Python

python
plt.figure(figsize=(6,10))
plt.imshow(img.crop((300,1000,2000,3000))); plt.axis('off')

(-0.5, 1699.5, 1999.5, -0.5)

Python

python
plt.figure(figsize=(6,8))
plt.imshow(img.crop((300,250,2000,950))); plt.axis('off')

(-0.5, 1699.5, 699.5, -0.5)

Python

python
plt.figure(figsize=(6,8))
plt.imshow(img.crop((300,600,2000,1300))); plt.axis('off')

(-0.5, 1699.5, 699.5, -0.5)

Python

python
from functools import lru_cache

def game_value(A, edges):
    n=len(A)
    tree=[[] for _ in range(n)]
    for u,v in edges:
        tree[u].append(v)
    # assume tree; root=0
    # state: current node u, removed set bitmask maybe
    from functools import lru_cache
    # We'll encode removed mask as tuple bools maybe limited; but exhaustive search quickly large.
    # For this small chain we can brute.
    n_nodes=len(A)
    @lru_cache(None)
    def val(u, removed_mask):
        # bitmask int of length n_nodes for nodes zeroed
        # if node value zeroed, its value=0.
        # It's Alice's turn.
        best=10**9
        # For each node i to zero next turn: consider all nodes 0..n-1
        for i in range(n_nodes):
            if removed_mask>>i & 1: # already zeroed
                continue
            new_mask=removed_mask | (1<<i)
            # compute Bob's best response
            # Bob chooses stop or child move
            # compute value of node u considering removal
            cur_val=0 if (new_mask>>u & 1) else A[u]
            best_child=-1
            # Bob stop
            best_score=cur_val
            # Bob move to each child
            for v in tree[u]:
                if (v==None): continue
                # if v is leaf: immediate value
                v_val=0 if (new_mask>>v & 1) else A[v]
                if tree[v]: # not leaf
                    v_val = val(v, new_mask)
                best_score=max(best_score, v_val)
            # Alice chooses min score
            best=min(best, best_score)
        return best
    return val(0,0)
# Test chain
A=[10,14,19]
edges=[(0,1),(1,2)]
print(game_value(A, edges))

Python

python
def get_value_chain(vals):
    A=vals
    edges=[(i,i+1) for i in range(len(A)-1)]
    return game_value(A, edges)
print(get_value_chain([10]))

Python

python
print(get_value_chain([10,5]))

Python

python
print(get_value_chain([25,5]))

Python

python
print(get_value_chain([1,100]))

Python

python
print(get_value_chain([50,60,70]))

Python

python
def game_value_tree(A, parent_edges):
    # edges list of (parent child) indexes
    n=len(A)
    tree=[[] for _ in range(n)]
    for u,v in parent_edges:
        tree[u].append(v)
    from functools import lru_cache
    @lru_cache(None)
    def val(u, mask):
        best=10**9
        for i in range(n):
            if mask>>i & 1: continue
            newmask=mask| (1<<i)
            cur_val = 0 if newmask>>u &1 else A[u]
            best_score=cur_val
            for v in tree[u]:
                v_val = 0 if newmask>>v &1 else A[v]
                if tree[v]:
                    v_val=val(v, newmask)
                best_score=max(best_score, v_val)
            best=min(best, best_score)
        return best
    return val(0,0)

A=[10,20,5,8]
edges=[(0,1),(0,2),(0,3)]
print(game_value_tree(A,edges))

Python

python
# verify proper enumeration for star tree manually by exploring step1 choices only maybe
from itertools import product

def star_game(A):
    n=len(A)
    children = [[1,2,3]]
    # simple star
    best=1e9
    for zero_index in range(n):
        # Step0 zero node
        newval=A.copy()
        newval[zero_index]=0
        # Bob choose stop or move to leaf
        stop_val=newval[0]
        move_vals=[newval[v] for v in children[0]]
        best_score=max(stop_val,max(move_vals))
        if best_score<best:
            best=best_score
            chosen_zero=zero_index
    print(best, chosen_zero)
star_game([10,20,5,8])

Python

python
def second_largest(lst):
    if len(lst)<2:
        return 0
    a=max(lst)
    b=max([x for x in lst if x!=a] or [a]) # if duplicates maybe.
    # If duplicates and same as max, second largest may equal same. Example [5,5] both duplicates -> second 5 as Alice zeros one of equal duplicates left behind; If all duplicates same, second largest = that same value.
    # We'll compute differently: Sort descending pick second.
    s=sorted(lst, reverse=True)
    return s[1] if len(s)>1 else 0

def compute_F(u, tree, A):
    # recursion
    if not tree[u]:
        # leaf but maybe root? But root leaf results list len1
        # We'll compute as per rule
        lst=[A[u]]
        return second_largest(lst)
    lst=[A[u]]
    for v in tree[u]:
        lst.append(compute_F(v,tree,A) if tree[v] else A[v])
    return second_largest(lst)

A=[10,20,5,8]
tree=[ [1,2,3], [], [], [] ]
print(compute_F(0,tree,A))

Python

python
A=[10,14,19]
tree=[ [1], [2], [] ]
print(compute_F(0,tree,A))

Python

python
A=[10]
tree=[ [] ]
print(compute_F(0,tree,A))

Python

python
A=[10,5]
tree=[[1], []]
print(compute_F(0,tree,A))

Python

python
A=[50,60,70]
tree=[[1],[2],[]]
print(compute_F(0,tree,A))

Python

python
A=[5,10,3,8]
tree=[[1,2],[ ],[3],[]]
print(compute_F(0,tree,A))

Python

python
A=[5,10,3,8]
edges=[(0,1),(0,2),(2,3)]
print(game_value_tree(A,edges))

Python

python
A=[5,5,5]
tree=[[1],[2],[]]
print(compute_F(0,tree,A))

Python

python
print(game_value_tree(A,[(0,1),(1,2)]))

Python

python
def second_max(arr):
    if len(arr)<2:
        return 0
    max1,max2=-10**18,-10**18
    for x in arr:
        if x>max1:
            max2=max1
            max1=x
        elif x>max2:
            max2=x
    return max2

def compute_S(root, adj, A):
    n=len(A)
    visited=[False]*n
    def dfs(u, p):
        candidates=[A[u]]
        for v in adj[u]:
            if v==p: continue
            child_val= A[v] if (len(adj[v])==1 and v!=root) else dfs(v,u)
            candidates.append(child_val)
        return second_max(candidates)
    return dfs(root,-1)

# test star
A=[10,20,5,8]
adj=[[1,2,3], [0],[0],[0]]
print(compute_S(0,adj,A))

Python

python
A=[10,14,19]
adj=[[1],[0,2],[1]]
print(compute_S(0,adj,A))

Python

python
A=[10]
adj=[[]]
print(compute_S(0,adj,A))

Python

python
A=[5,10,3,8]
adj=[[1,2],[0],[0,3],[2]]
print(compute_S(0,adj,A))

Python

python
A=[5,5,5]
adj=[[1],[0,2],[1]]
print(compute_S(0,adj,A))

Python

python
A=[5,5,5]
adj=[[1],[0,2],[1]]
print(compute_S(0,adj,A))

Pensamiento

A=[5,5,5] adj=[[1],[0,2],[1]] print(compute_S(0,adj,A))

Thought for 7m 5s

Respuesta

I misunderstood the rules last time – Alice is allowed to set one new node to 0 every time Bob is about to move (not just once at the beginning).
With that extra power the game at a node u works like this, assuming it is Alice’s turn and Bob is standing on u.

  • Bob’s possible scores this turn
Bob’s actionScore he would get
stop hereA[u]
move to child v• if v is a leaf → A[v] (game ends immediately)<br>• otherwise → value we’ll call S[v]

(S[v] is the optimal score Bob can secure from that child subtree when it is Alice’s turn again.)

  • Alice’s move – she can erase the value of exactly one node before Bob chooses.
    Whatever single option in the above table is currently best for Bob, Alice will erase it.
    Bob then takes the best of the remaining options.

Consequently, at every non-leaf node the value of the game is simply the second-largest number in the options table.

Recurrence

Let

text
value(u → v) =  A[v]        if v is a leaf
             =  S[v]        otherwise

Create a list

choices = { A[u] } ∪ { value(u → v) | v is a child of u }

Then

S[u] = second_largest(choices)   // if choices has only one element → 0

A true leaf has no children, so choices = {A[leaf]} and S[leaf] = 0.

Evaluating the tree bottom-up gives S[root], the final score when both play optimally.

The recurrence is O(N) because each edge is visited once and “second-largest” is computed in O(1).

Correct C++ implementation

cpp
#include <bits/stdc++.h>
using namespace std;
using ll = long long;

const ll NEG_INF = LLONG_MIN;

struct Tree {
    vector<vector<int>> g;
    vector<ll> A;
    int N;
    Tree(int n): g(n), A(n), N(n) {}
    void addEdge(int u, int v) {
        g[u].push_back(v);
        g[v].push_back(u);
    }
    // returns the game value S[u] when it is Alice's turn and Bob stands on u
    ll solve() {
        return dfs(0, -1);
    }
private:
    ll dfs(int u, int p) {
        // gather Bob's possible scores from this node
        ll max1 = NEG_INF, max2 = NEG_INF;   // largest and 2nd-largest
        auto upd = [&](ll x) {
            if (x > max1) { max2 = max1; max1 = x; }
            else if (x > max2) { max2 = x; }
        };

        upd(A[u]);                 // stopping here

        bool isLeaf = true;
        for (int v: g[u]) if (v != p) {
            isLeaf = false;
            ll childScore;
            if (g[v].size() == 1)          // v is a leaf (only parent)
                childScore = A[v];
            else                           // internal – recurse
                childScore = dfs(v, u);
            upd(childScore);
        }
        // If u is a (true) leaf, max2 is still NEG_INF – score is 0.
        return (max2 == NEG_INF ? 0 : max2);
    }
};

int main() {
    ios::sync_with_stdio(false);
    cin.tie(nullptr);

    int N, M, Two;
    if (!(cin >> N)) return 0;
    cin >> M >> Two;          // Two is always 2 – not used

    Tree T(N);
    for (ll &x : T.A) cin >> x;    // node values, 1-based in statement

    for (int i = 0; i < M; ++i) {
        int u, v;
        cin >> u >> v;
        --u; --v;                  // convert to 0-based
        T.addEdge(u, v);
    }

    cout << T.solve() << '\n';
    return 0;
}

Why it matches the sample (N=3, A=[10,14,19])

  • At node 3 (leaf) → S[3] = 0
  • At node 2: choices = {14 (stop), 19 (move)} → second-largest = 14 → S[2]=14
  • At node 1: choices = {10 (stop), 14 (move)} → second-largest = 10 → score = 10

Exactly the required output.

This single, bottom-up pass handles every tree in O(N) time and O(N) memory.

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