Overview

This is a problem of finding the total number of petals falling from cherry trees within the interval you pass through when traveling the shortest distance from point \(S\) to point \(T\).

Analysis

Key Observations

1. Traveling the shortest distance means moving in a straight line from \(S\) to \(T\). In other words, you only pass through points between \(S\) and \(T\).

2. For example, if \(S = 2\) and \(T = 5\), the points you pass through are \(2, 3, 4, 5\). If \(S = 5\) and \(T = 2\), the points you pass through are still \(2, 3, 4, 5\) (just in the opposite direction).

3. Therefore, we only need to count the petals from cherry trees in the interval from \(\min(S, T)\) to \(\max(S, T)\).

Problem with the Naive Approach

A naive approach of “checking all points from \(1\) to \(N\)” would result in \(O(N)\) time complexity, causing TLE (Time Limit Exceeded) since \(N\) can be as large as \(10^9\).

Solution

While \(N\) is large, the number of cherry trees \(M\) is at most \(2 \times 10^5\), which is small. Instead of checking all points, we only need to check the points where cherry trees exist.

Algorithm

1. Determine the travel interval \([\text{left}, \text{right}]\) from the start point \(S\) and goal point \(T\)

   • \(\text{left} = \min(S, T)\)
   • \(\text{right} = \max(S, T)\)

2. For each cherry tree, check whether its position \(P_i\) is within the travel interval

   • If \(\text{left} \leq P_i \leq \text{right}\), add the number of petals \(V_i\) to the total

3. Output the total

Concrete Example

N=10, M=3, S=3, T=7
Cherry trees: (P=2, V=100), (P=5, V=200), (P=8, V=300)

The travel interval is \([3, 7]\). - \(P=2\): Outside the interval (\(2 < 3\)) → Do not count - \(P=5\): Inside the interval (\(3 \leq 5 \leq 7\)) → Add \(200\) - \(P=8\): Outside the interval (\(8 > 7\)) → Do not count

Answer: \(200\)

Complexity

• Time Complexity: \(O(M)\)
  • For each of the \(M\) cherry trees, we determine whether it’s in the interval in \(O(1)\)
• Space Complexity: \(O(1)\)
  • We process while reading without storing the cherry tree information, requiring only constant space

Implementation Notes

1. Determining interval endpoints: Since the problem statement does not specify the relative order of \(S\) and \(T\), we need to normalize the interval using min and max.

2. Overflow prevention: Since \(V_i\) can be at most \(10^9\) and \(M\) can be at most \(2 \times 10^5\), the total can reach approximately \(2 \times 10^{14}\). In Python, you don’t need to worry about integer overflow, but in C++ and similar languages, you need to use long long.

3. Efficient input processing: There’s no need to store all cherry tree information in an array and sort it. By checking and adding while reading, we can improve space efficiency.

Source Code

def main():
    N, M = map(int, input().split())
    S, T = map(int, input().split())
    
    # Determine the travel interval
    left = min(S, T)
    right = max(S, T)
    
    total = 0
    for _ in range(M):
        P, V = map(int, input().split())
        # Check if the cherry tree is within the travel interval
        if left <= P <= right:
            total += V
    
    print(total)

if __name__ == "__main__":
    main()

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