发现这个有趣的问题Dynamic Planning Approach for Radio Tower Placement in Cities

在 ByteLand 中，沿轴有 n(≤ 105) 个城市，第 i 个城市位于 (A₁,0)（对于 1≤ i ≤ n，A≤ 10°）。 在ByteLand，有一家制造无线电塔的电信公司。每个塔的覆盖半径为 d，即，当且仅当 x - y ≤ d 时，(y, 0) 处的塔可覆盖 (2,0) 处的城市。提示：1.目标解决方案基于动态规划，2.问题陈述本身包含有关如何定义 DP 函数的提示 该公司已确定 m(≤1200) 个可能放置无线电塔的位置。这些位置中的第 j 个位于 (B3, 0)（B；≤ 10°，对于 1≤ j ≤ m）。该公司只能在这些地点放置无线电塔。如果一个城市有多个塔提供服务，该公司也会面临困难。也就是说，假设有两个塔T1、T2使得某个城市A；位于两座塔的覆盖范围内。在这种情况下，A 中的设备将无法决定与 T1、T2 中的哪个塔进行通信。为了避免这个问题，他们想建造塔，这样这样的情况就永远不会发生 发生。 因此，该公司希望放置一些塔楼，以便 (i) 每个城市至少被一座塔楼覆盖，并且 (ii) 没有一个城市被超过一座塔楼覆盖。为了帮助他们找到这样的作业，请解决以下问题。 对于每个 1≤ k ≤ m 的整数 k，计算将无线电塔放置在 m 个给定位置中的 k 个位置的方法数，以便每个城市都被一个塔覆盖。由于答案可能很大，所以输出它对 109+7 取模，即，如果答案是 x，则打印 z 除以 109+7 时的余数。 暗示。您可能想要解决以下（看似更难）问题：对于每个整数 1≤ i ≤ m，且 1 ≤ j ≤ m，我们可以通过多少种方式选择塔位置，以便 (i) 最右边的塔位于 B[i ], (ii) B[1], B[2],... B[i] 中恰好选择了 j 个塔（包括 B[i]），并且 (iii) B[i] 左边的每个城市是唯一被塔覆盖的吗？看看你是否可以使用这些问题将原始问题简化为更小的子问题。示例解决方案的复杂度为 O(m² + n) 或 O(m² logn+n)。如果您的解决方案需要 (m³) 时间，那么您的期望分数将为 3.6/6.0。 输入输出 每个输入文件包含多个测试实例。第一行包含一个整数 C，表示测试实例的数量。 C 实例的描述如下。 每个测试实例由三行组成： • 第一行包含三个空格分隔的整数n、m、d。 • 第二行包含n 个空格分隔的整数A1, A2,... An。 • 第三行包含 m 个空格分隔的整数 B1, B2,... Bm • 保证城市和潜在塔位置按升序给出，即Ai > Ai-1 和B； > Bj-1 对于每个 1 <i<n 且 1<j≤m 成立。输出 m 个空格分隔的整数，其中第 i 个整数表示精确选择 i 个塔来唯一覆盖所有城市的方法数。 在此作业中，您将获得包含输入和输出代码的示例提交。您只需要完成一个函数，给定 d 以及城市和塔的位置，返回 m 个整数的数组。

5 5 2
1 2 3 4 5
1 2 3 4 5

0 2 0 0 0

说明：有两种方法可以覆盖城市。第一种方法是在 {1,4} 放置塔，第二种方法是在 {2,5} 放置塔。第一种方式中，A[1]、A[2] 处的城市仅被 B[1] 处的塔覆盖，而 A[3]、A[4]、A[5] 处的城市仅被 B[1] 处的塔覆盖。 B[5]。两种方式都需要 2 个塔。因此答案是{0,2,0,0,0}。

MOD = 10 ** 9 + 7

def compute_cost(A, B, i, d):
    left = B[i] - d + 1
    right = B[i] + d - 1
    count = 0

    for city in A:
        if left <= city <= right:
            count += 1  # This city is covered by this tower

    return count

def solve_coverage(n, m, d, A, B):
    dp = [[float('inf')] * (m + 1) for _ in range(m + 1)]
    count_towers = [[0] * (m + 1) for _ in range(m + 1)]

    for i in range(m + 1):
        dp[0][i] = 0
        count_towers[0][i] = 1

    for t in range(1, m + 1):
        for i in range(1, m + 1):
            cost = compute_cost(A, B, i - 1, d)
            dp[t][i] = cost + dp[t - 1][i - 1]
            count_towers[t][i] = count_towers[t - 1][i - 1]

            for j in range(i - 1, 0, -1):
                cost = compute_cost(A, B, i - 1, d) - compute_cost(A, B, j - 1, d)
                new_cost = dp[t - 1][j - 1] + cost

                if new_cost < dp[t][i]:
                    dp[t][i] = new_cost
                    count_towers[t][i] = count_towers[t - 1][j - 1]
                elif new_cost == dp[t][i]:
                    count_towers[t][i] = (count_towers[t][i] + count_towers[t - 1][j - 1]) % MOD

    result = [0] * m
    for t in range(1, m + 1):
        result[t - 1] = dp[t][m] % MOD

    return result

def main():
    import sys
    if len(sys.argv) != 2:
        print("Usage: python telecom_towers.py <inputfile>")
        return

    inputfile = sys.argv[1]

    with open(inputfile, 'r') as file:
        data = file.read().split()

    index = 0
    C = int(data[index])
    index += 1
    results = []

    for _ in range(C):
        n = int(data[index])
        m = int(data[index + 1])
        d = int(data[index + 2])
        index += 3
        A = list(map(int, data[index:index + n]))
        index += n
        B = list(map(int, data[index:index + m]))
        index += m

        result = solve_coverage(n, m, d, A, B)
        results.append(" ".join(map(str, result)))

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

MOD = 10**9 + 7

def solve_coverage(n, m, d, A, B):
    """
    Calculates the number of ways to place radio towers in k positions out of m
    to cover all cities uniquely.

    Args:
        n: The number of cities.
        m: The number of potential tower locations.
        d: The coverage radius of each tower.
        A: A list of city locations.
        B: A list of potential tower locations.

    Returns:
        A list of m integers, where the i-th integer represents the number of
        ways to choose i towers to uniquely cover all cities.
    """

    # dp[i][j] stores the number of ways to choose j towers from the first i
    # potential locations, such that all cities up to the rightmost covered
    # city by these j towers are uniquely covered.
    dp = [[0 for _ in range(m + 1)] for _ in range(m + 1)]

    # Base case: no towers placed.
    for i in range(m + 1):
        dp[i][0] = 1

    # Iterate over potential tower locations.
    for i in range(1, m + 1):
        # Iterate over the number of towers to be placed.
        for j in range(1, i + 1):
            # Case 1: Don't place the current tower.
            dp[i][j] = dp[i - 1][j]

            # Case 2: Place the current tower.
            # Find the rightmost city covered by the current tower.
            rightmost_city = -1
            for city_index in range(n):
                if A[city_index] <= B[i - 1] + d:
                    rightmost_city = city_index

            # Find the leftmost city that can be covered by the current tower
            # without overlapping with previously placed towers.
            leftmost_city = 0
            for k in range(i - 1, 0, -1):
                if B[k - 1] + d >= A[leftmost_city]:
                    leftmost_city = k

            # Add the number of ways to place j - 1 towers from the leftmost
            # city to the current tower location.
            dp[i][j] = (dp[i][j] + dp[leftmost_city][j - 1]) % MOD

    # The result array stores the number of ways to choose k towers.
    result = [0] * m
    for k in range(1, m + 1):
        result[k - 1] = dp[m][k]

    return result

def main():
    """
    Reads input from a file, solves the tower placement problem, and prints
    the results.
    """
    import sys
    if len(sys.argv) != 2:
        print("Usage: python telecom_towers.py <inputfile>")
        return

    inputfile = sys.argv[1]

    with open(inputfile, 'r') as file:
        data = file.read().split()

    index = 0
    C = int(data[index])
    index += 1
    results = []

    for _ in range(C):
        n = int(data[index])
        m = int(data[index + 1])
        d = int(data[index + 2])
        index += 3
        A = list(map(int, data[index:index + n]))
        index += n
        B = list(map(int, data[index:index + m]))
        index += m

        result = solve_coverage(n, m, d, A, B)
        results.append(" ".join(map(str, result)))

    for result in results:
        print(result)

if __name__ == "__main__":
    main()

Explanation:

The solution uses dynamic programming to solve the problem efficiently.

1. State Definition:

2. dp[i][j] represents the number of ways to choose j towers from the first i potential tower locations ( B[0] to B[i-1] ) such that all cities up to the rightmost city covered by these j towers are uniquely covered.

3. Base Case:

4. dp[i][0] = 1 for all i . This means there's one way to choose zero towers (i.e., not placing any tower).

5. Recurrence Relation:

6. For each potential tower location i and number of towers j :

   • Case 1: Don't place the tower at location i . In this case, the number of ways to cover cities remains the same as if we considered only the first i-1 tower locations: dp[i][j] = dp[i-1][j] .

   • Case 2: Place the tower at location i .

   • Find the rightmost city rightmost_city covered by the tower at B[i-1] .
   • Find the leftmost city leftmost_city that can be covered by the tower at B[i-1] without overlapping coverage from previously placed towers. We need to ensure that no city is covered by more than one tower.
   • To avoid overlapping coverage, we only consider placing towers from leftmost_city to i-1 and add the number of ways to place j-1 towers in this range: dp[i][j] = (dp[i][j] + dp[leftmost_city][j - 1]) % MOD .

7. Final Result:

8. The final result array result stores the number of ways to choose k towers for each k from 1 to m . We get this from dp[m][k] , which represents choosing k towers from all m locations.

Time and Space Complexity:

• Time Complexity: O(m² * n), where m is the number of potential tower locations and n is the number of cities. This is because we have nested loops iterating over m , j , and sometimes n (while finding rightmost_city and leftmost_city ).
• Space Complexity: O(m²), as we are using a 2D DP table of size (m+1) * (m+1).

Improvements:

• The time complexity can be improved to O(m² + n) or O(m² log n + n) by using binary search or other efficient data structures to find rightmost_city and leftmost_city more efficiently.