제출 #1271695

#	제출 시각	아이디	문제	언어	결과	실행 시간	메모리
1271695		lazypanda5050	Rabbit Carrot (LMIO19_triusis)	C++20	14 / 100	1100 ms	98376 KiB

triusis

#include <iostream>
#include <vector>
#include <algorithm>
#include <climits>

using namespace std;

int main() {
    int N, M;
    cin >> N >> M;
    
    vector<int> heights(N);
    for (int i = 0; i < N; i++) {
        cin >> heights[i];
    }
    
    // dp[i][mods] = maximum height we can reach at pole i using exactly 'mods' modifications
    vector<vector<int>> dp(N + 1, vector<int>(N + 1, -1));
    
    // Base case: we start at height 0 before any pole
    dp[0][0] = 0;
    
    for (int i = 1; i <= N; i++) {
        int currentHeight = heights[i - 1];
        
        for (int mods = 0; mods <= min(i, N); mods++) {
            // Option 1: Don't modify this pole
            for (int prevMods = 0; prevMods <= mods && prevMods < i; prevMods++) {
                if (dp[i - 1][prevMods] != -1) {
                    int prevHeight = dp[i - 1][prevMods];
                    
                    // Can we jump to currentHeight from prevHeight?
                    if (currentHeight <= prevHeight + M) { // Can jump up at most M, down any amount
                        dp[i][mods] = max(dp[i][mods], currentHeight);
                    }
                }
            }
            
            // Option 2: Modify this pole (use one more modification)
            if (mods > 0) {
                for (int prevMods = 0; prevMods < mods && prevMods < i; prevMods++) {
                    if (dp[i - 1][prevMods] != -1) {
                        int prevHeight = dp[i - 1][prevMods];
                        
                        // If we modify this pole, we can set it to any height from 0 to prevHeight + M
                        // Best strategy: set it to prevHeight + M to maximize future reach
                        int optimalHeight = prevHeight + M;
                        dp[i][mods] = max(dp[i][mods], optimalHeight);
                    }
                }
            }
        }
    }
    
    // Find the minimum number of modifications needed
    for (int mods = 0; mods <= N; mods++) {
        if (dp[N][mods] != -1) {
            cout << mods << endl;
            return 0;
        }
    }
    
    // Should never reach here if input is valid
    cout << N << endl;
    return 0;
}

• Subtask 1: 14 / 14
• Subtask 2: 0 / 21
• Subtask 3: 0 / 28
• Subtask 4: 0 / 37