Overview

This is a problem of finding the shortest path that visits all intermediate waypoints from a start point to reach a goal, which is a type of Traveling Salesman Problem (TSP). We solve it efficiently using bitmask DP.

Analysis

Key Observations

1. Shortest distance on a grid is Manhattan distance: The shortest distance between two points on a grid without obstacles can be calculated using Manhattan distance (\(|r_1 - r_2| + |c_1 - c_2|\)). There’s no need to use BFS or similar algorithms.

2. The order of visiting waypoints matters: The total travel distance changes depending on the order in which we visit the \(M\) waypoints. We need to find the optimal order.

Problem with the Naive Approach

A method that tries all visiting orders of waypoints would need to examine \(M!\) permutations. For \(M = 15\), \(15! \approx 1.3 \times 10^{12}\), which cannot be processed in time.

Solution

We use bitmask DP. We manage “which waypoints have been visited” using a bitmask, avoiding recalculation of the same states. This reduces the complexity from \(O(M! )\) to \(O(2^M \times M^2)\).

Algorithm

1. Coordinate Conversion and Manhattan Distance Calculation

Conversion from cell number \(k\) (1-indexed) to coordinates: - Row: \((k-1) \div N\) - Column: \((k-1) \mod N\)

2. Building the Distance Matrix

Precompute the Manhattan distances between all \((M+2)\) points: start \(S\), goal \(T\), and waypoints \(P_1, \ldots, P_M\).

3. Bitmask DP

Define the state as follows: - \(dp[\text{mask}][i]\) = minimum travel distance when starting from the start, having visited the set of waypoints represented by \(\text{mask}\), and currently at waypoint \(i\)

Initial state: Direct movement from start to each waypoint \(i\) $\(dp[2^i][i] = \text{dist}(S, P_i)\)$

Transition: Move from waypoint \(\text{last}\) to an unvisited waypoint \(\text{next}\) $\(dp[\text{mask} | 2^{\text{next}}][\text{next}] = \min(dp[\text{mask} | 2^{\text{next}}][\text{next}], dp[\text{mask}][\text{last}] + \text{dist}(P_{\text{last}}, P_{\text{next}}))\)$

4. Computing the Answer

After visiting all waypoints (\(\text{mask} = 2^M - 1\)), the answer is the minimum value obtained by adding the distance from each final waypoint to the goal: $\(\text{ans} = \min_{i=0}^{M-1} \left( dp[2^M - 1][i] + \text{dist}(P_i, T) \right)\)$

Complexity

• Time complexity: \(O(M^2 + 2^M \times M^2)\)
  • Building the distance matrix: \(O((M+2)^2) = O(M^2)\)
  • Bitmask DP: Number of states \(O(2^M \times M)\), transitions from each state \(O(M)\)
• Space complexity: \(O(2^M \times M)\)
  • Size of the DP table

Since \(M \leq 15\), approximately \(2^{15} \times 15^2 \approx 7 \times 10^6\) operations, which runs fast enough.

Implementation Notes

• Index management: In the points array, \(S\) is placed at index 0, \(T\) at index 1, and waypoints from index 2 onwards. Since waypoints are managed as 0 to \(M-1\) in the DP, accessing the points array requires adding +2.
• Case when \(M = 0\): If there are no waypoints, the answer is simply the Manhattan distance from \(S\) to \(T\).
• Bitmask operations: mask & (1 << i) checks if the \(i\)-th waypoint has been visited, mask | (1 << i) updates the \(i\)-th waypoint as visited.

Source Code

import sys
from itertools import permutations

def solve():
    input_data = sys.stdin.read().split()
    idx = 0
    N = int(input_data[idx]); idx += 1
    M = int(input_data[idx]); idx += 1
    S = int(input_data[idx]); idx += 1
    T = int(input_data[idx]); idx += 1
    
    P = []
    for i in range(M):
        P.append(int(input_data[idx])); idx += 1
    
    # Convert cell number to coordinates (row, col) (0-indexed)
    def to_coord(num):
        num -= 1  # 0-indexed
        row = num // N
        col = num % N
        return (row, col)
    
    # Calculate Manhattan distance
    def manhattan(p1, p2):
        r1, c1 = to_coord(p1)
        r2, c2 = to_coord(p2)
        return abs(r1 - r2) + abs(c1 - c2)
    
    # All important points (S, T, P[0..M-1])
    # S: index 0
    # T: index 1
    # P[i]: index i+2
    points = [S, T] + P
    num_points = len(points)
    
    # Calculate distance matrix
    dist = [[0] * num_points for _ in range(num_points)]
    for i in range(num_points):
        for j in range(num_points):
            dist[i][j] = manhattan(points[i], points[j])
    
    if M == 0:
        # No waypoints, go directly from S to T
        print(dist[0][1])
        return
    
    # Solve TSP using bitmask DP
    # dp[mask][i] = minimum distance when starting from S, having visited waypoints represented by mask, and currently at waypoint i
    # Waypoint indices are 0 to M-1 (indices 2 to M+1 in the points array)
    
    INF = float('inf')
    dp = [[INF] * M for _ in range(1 << M)]
    
    # Initial state: distance from S to each waypoint
    for i in range(M):
        dp[1 << i][i] = dist[0][i + 2]  # From S to waypoint i
    
    # DP transition
    for mask in range(1, 1 << M):
        for last in range(M):
            if not (mask & (1 << last)):
                continue
            if dp[mask][last] == INF:
                continue
            # Move to next waypoint
            for next_p in range(M):
                if mask & (1 << next_p):
                    continue
                new_mask = mask | (1 << next_p)
                new_dist = dp[mask][last] + dist[last + 2][next_p + 2]
                if new_dist < dp[new_mask][next_p]:
                    dp[new_mask][next_p] = new_dist
    
    # After visiting all waypoints, move to T
    full_mask = (1 << M) - 1
    ans = INF
    for last in range(M):
        if dp[full_mask][last] < INF:
            total = dp[full_mask][last] + dist[last + 2][1]  # From last waypoint to T
            ans = min(ans, total)
    
    print(ans)

solve()

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