Overview

This problem asks whether we can reach the goal from the starting point by moving by \(+1\) or \(+2\) at each step, while enduring the stamina depletion at each location. If it is possible to reach the goal, we want to find the maximum possible final stamina. We can solve this using DP (Dynamic Programming) to optimize the path selection and the timing of using food.

Analysis

It is optimal to eat food immediately

The most crucial observation in this problem is that it is always optimal to eat any food immediately upon obtaining it.

The reasons are as follows:

• There is no upper limit on stamina (it can exceed the initial value \(K\)).
• Whether you carry the food over or eat it immediately, the total amount of food you can eat later remains the same.
• The retirement check is based on whether the stamina drops below \(0\) immediately after depletion.

In other words, the earlier you eat the food, the higher your stamina will be, making it easier to pass subsequent retirement checks. Conversely, carrying it over and delaying consumption offers absolutely no benefit. Therefore, we can safely fix the strategy to “eat all food the moment it is obtained.”

With this observation, the exponential choice of “which food to eat and when” disappears, and the only thing we need to consider is the path (whether to move \(+1\) or \(+2\)).

Considering the state at each location

Let \(dp[i]\) be the stamina after arriving at location \(i\) and completing all processing (depletion \(\to\) check \(\to\) acquisition \(\to\) eating).

• If the stamina just before arriving at location \(i\) (i.e., the stamina after processing at the previous location) is \(s\), then after depletion, it becomes \(s - H_i\).
• The condition to not retire is \(s - H_i \geq 0\), which is \(s \geq H_i\).
• If we can pass through, we obtain and immediately eat the food \(food_i\), resulting in \(dp[i] = s - H_i + food_i\).

Since we can arrive at location \(i\) from either location \(i-1\) or \(i-2\), the candidates for \(s\) are \(dp[i-1]\) and \(dp[i-2]\).

Issues with the naive approach

If we enumerate all paths, the number of paths becomes exponential. Furthermore, searching all \(2^M\) combinations of “which food to eat” would be far too slow to run within the time limit. Since \(N\) can be up to \(10^6\), we need a linear-time DP. We can solve this by combining the “eat immediately” observation with DP.

Algorithm

Let \(dp[i]\) be the “maximum stamina immediately after completing the processing at location \(i\)”. If location \(i\) is unreachable, \(dp[i]\) is undefined (represented as \(-1\) in the code).

Initialization (Location \(1\))

\[ dp[1] = \begin{cases} K - H_1 + food_1 & (K - H_1 \geq 0) \\ \text{unreachable} & (\text{otherwise}) \end{cases} \]

Transitions (\(i \geq 2\))

Among locations \(i-1\) and \(i-2\), let \(best\) be the maximum value of \(dp[\cdot]\) that is reachable and satisfies \(dp[\cdot] \geq H_i\).

\[ dp[i] = best - H_i + food_i \quad (\text{when } best \geq H_i) \]

If no such location exists, location \(i\) is unreachable.

The reason we can greedily choose the maximum \(best\) here is to maximize the final stamina, and having “more stamina” is always advantageous for subsequent checks.

Answer

If \(dp[N]\) is reachable, output its value; otherwise, output \(-1\).

Concrete Example

For example, suppose before location \(i\), we have \(dp[i-1] = 5\), \(dp[i-2] = 8\), and \(H_i = 6\). Since \(dp[i-1] = 5 < 6\), we cannot endure the depletion from \(i-1\) and will retire. On the other hand, since \(dp[i-2] = 8 \geq 6\), we can pass through if we come from \(i-2\), resulting in \(dp[i] = 8 - 6 + food_i\). In this way, we choose the candidate that is larger and satisfies the condition.

Complexity

• Time Complexity: \(O(N + M)\)
  • The DP visits each location only once, taking \(O(N)\) time, and reading the food items takes \(O(M)\) time.
• Space Complexity: \(O(N)\)
  • To store the arrays dp, H, and food.

Implementation Details

• Any reachable \(dp\) value is guaranteed to be at least \(0\). Therefore, if we represent “unreachable” with \(-1\) (NEG), since \(H_i \geq 0\), an unreachable cell will never falsely satisfy the transition condition dp[pred] >= H_i. This allows us to perform safe checks without any special conditional branches.

• Since \(N\) can be as large as \(10^6\), we should read the input in bulk using sys.stdin.buffer.read() for fast I/O processing.

• Since there is at most one food item per location, it is convenient to map locations to their recovery amounts using an array like food[p] = r.

  Source Code

import sys

def main():
    data = sys.stdin.buffer.read().split()
    idx = 0
    N = int(data[idx]); idx += 1
    M = int(data[idx]); idx += 1
    K = int(data[idx]); idx += 1
    H = [0] * (N + 1)
    for i in range(1, N + 1):
        H[i] = int(data[idx]); idx += 1
    food = [0] * (N + 1)
    for _ in range(M):
        p = int(data[idx]); idx += 1
        r = int(data[idx]); idx += 1
        food[p] = r

    NEG = -1  # unreachable (reachable dp values are always >= 0)
    dp = [NEG] * (N + 1)

    if K - H[1] >= 0:
        dp[1] = K - H[1] + food[1]

    for i in range(2, N + 1):
        hi = H[i]
        best = NEG
        # from i-1
        v = dp[i - 1]
        if v >= hi:
            best = v
        # from i-2
        if i - 2 >= 1:
            v = dp[i - 2]
            if v >= hi and v > best:
                best = v
        if best >= hi:  # best != NEG and best >= hi (hi>=0 so NEG=-1 fails)
            dp[i] = best - hi + food[i]

    print(dp[N] if dp[N] >= 0 else -1)

main()

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