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question:Alice is organizing a coding event and wants to ensure a fair distribution of prizes. She has a list of `n` participants and each participant is initially assigned a score of `0`. During the event, a total of `m` operations are performed, where each operation can either increase or decrease the score of a subset of participants by a particular value. Alice wants to know the final scores of all participants after all operations are executed. Each operation is given in the form of three integers `a`, `b`, and `v`, meaning that the score of each participant from index `a` to index `b` (inclusive) is increased by `v`. Write a program that takes the number of participants and the number of operations as its inputs and then processes each operation to update the scores of the participants accordingly. Finally, the program should output the final scores of all participants. # Input - The first line contains two integers `n` and `m` (`1 <= n, m <= 1000`) — the number of participants and the number of operations. - The next `m` lines each contain three integers `a`, `b`, and `v` (`1 <= a, b <= n`, `-1000 <= v <= 1000`, `a <= b`), representing an operation. # Output - Print `n` integers separated by spaces — the final scores of all participants. # Example Input ``` 5 3 1 3 2 2 5 -1 3 4 3 ``` Output ``` 2 1 4 2 -1 ``` Explanation - Initially, the scores of all participants are `[0, 0, 0, 0, 0]`. - After the first operation (1, 3, 2), the scores become `[2, 2, 2, 0, 0]`. - After the second operation (2, 5, -1), the scores become `[2, 1, 1, -1, -1]`. - After the third operation (3, 4, 3), the scores become `[2, 1, 4, 2, -1]`.

answer:def calculate_final_scores(n, m, operations): Calculate final scores after applying a series of operations. :param n: Number of participants :param m: Number of operations :param operations: List of operations where each operation is a tuple (a, b, v) :return: List of final scores of all participants scores = [0] * n for operation in operations: a, b, v = operation for i in range(a - 1, b): scores[i] += v return scores

question:Sophia loves gardening and has a beautiful garden with paths laid out in a grid. Each cell in the grid can either contain a plant or be empty. The grid is represented as an `n x m` matrix where each cell contains either a 0 (empty) or a 1 (plant). Sophia wants to water her plants, but she can only water them in rectangular sections due to the shape of her watering device. She can choose to water any rectangular section of the grid such that it contains only plants (1s) and no empty cells (0s). Help Sophia determine the maximum possible area of the rectangular section she can water, which contains only plants. The first line of input contains two integers `n` and `m` (1 ≤ n, m ≤ 1000) — the dimensions of the garden grid. Each of the next `n` lines contains `m` integers (each either 0 or 1), representing the grid. Print a single integer — the maximum area of the rectangular section that can be watered. Example: Input: 4 5 1 0 1 0 0 1 0 1 1 1 1 1 1 1 1 1 0 0 1 0 Output: 6 Explanation: The largest rectangle containing only 1s has an area of 6 (from rows 2-3 and columns 3-5).

answer:def max_area_of_plant_section(n, m, grid): if not grid: return 0 max_area = 0 heights = [0] * m for row in grid: for i in range(m): if row[i] == 0: heights[i] = 0 else: heights[i] += 1 max_area = max(max_area, largest_rectangle_area(heights)) return max_area def largest_rectangle_area(heights): stack = [] max_area = 0 heights.append(0) for i in range(len(heights)): while stack and heights[i] < heights[stack[-1]]: h = heights[stack.pop()] w = i if not stack else i - stack[-1] - 1 max_area = max(max_area, h * w) stack.append(i) heights.pop() return max_area # Example usage: # n = 4 # m = 5 # grid = [ # [1, 0, 1, 0, 0], # [1, 0, 1, 1, 1], # [1, 1, 1, 1, 1], # [1, 0, 0, 1, 0] # ] # print(max_area_of_plant_section(n, m, grid)) # Output: 6

question:You are given a directed graph represented as a list of edges. Each edge connects two distinct vertices and has a non-negative integer weight. Your task is to find the shortest path from a given starting vertex to a given target vertex. If no such path exists, return -1. The first line contains three integers n, m, and k — the number of vertices (2 ≤ n ≤ 1000), the number of edges (0 ≤ m ≤ 5000), and the number of queries (1 ≤ k ≤ 100). Each of the next m lines contains three integers u, v, and w indicating there is a directed edge from vertex u to vertex v with weight w (1 ≤ w ≤ 1000). Each of the next k lines contains two integers a and b — representing a query where you need to find the shortest path from vertex a to vertex b. Vertices are numbered from 1 to n. Print k lines, one for each query. For each query, print the length of the shortest path from a to b, or -1 if no path exists.

answer:import heapq import sys def dijkstra(n, edges, src): Perform Dijkstra's algorithm to find the shortest paths from src to all other vertices in a graph with n vertices. graph = {i: [] for i in range(1, n + 1)} for u, v, w in edges: graph[u].append((v, w)) dist = {i: float('inf') for i in range(1, n + 1)} dist[src] = 0 pq = [(0, src)] # (distance, vertex) while pq: current_dist, u = heapq.heappop(pq) if current_dist > dist[u]: continue for v, weight in graph[u]: distance = current_dist + weight if distance < dist[v]: dist[v] = distance heapq.heappush(pq, (distance, v)) return dist def shortest_paths(n, edges, queries): results = [] for start, end in queries: distances = dijkstra(n, edges, start) if distances[end] == float('inf'): results.append(-1) else: results.append(distances[end]) return results

question:A research team is investigating a novel algorithm to distribute water efficiently among different sections of a large agricultural field. The field is divided into multiple sections, each represented as a node in a tree structure, where the edges represent direct water pipes between the sections. The objective is to ensure that each section receives exactly one unit of water. Unfortunately, due to the topography, there is a time interval assigned to each node, indicating the amount of time it takes for water to travel from the water source (root) to that section. The team needs to determine the minimum total time required to distribute water to all sections of the field if the water can only travel from one section to another through the pipes given by the tree structure. The input consists of: - The first line contains an integer n (2 ≤ n ≤ 100000) — the number of sections (nodes) in the field. - The next n-1 lines each contain two integers u and v (1 ≤ u, v ≤ n, u ≠ v) — indicating there is a direct pipe (edge) between sections u and v. The water source is always at node 1. For each section (node) i (1 ≤ i ≤ n), you need to calculate the time it takes for water to travel from the water source to that section and then determine the maximum of these travel times, which will be the total time required to distribute water to all sections. Print a single integer — the minimum total time required to ensure all sections receive exactly one unit of water. # Example **Input:** ``` 5 1 2 1 3 3 4 3 5 ``` **Output:** ``` 2 ``` Note: The output '2' is calculated as the maximum distance from node 1 to any other node in the tree. In this example, the travel time to nodes 2 and 3 is 1, and to nodes 4 and 5 is 2. Therefore, the minimum time required to ensure that every section gets water is 2.

answer:from collections import deque def min_total_time(n, edges): Returns the minimum total time required to ensure all sections receive exactly one unit of water. def bfs(start_node): Breadth-First Search to calculate the maximum distance from the start node distances = [-1] * (n + 1) distances[start_node] = 0 queue = deque([start_node]) max_distance = 0 while queue: current = queue.popleft() for neighbor in graph[current]: if distances[neighbor] == -1: distances[neighbor] = distances[current] + 1 queue.append(neighbor) max_distance = max(max_distance, distances[neighbor]) return max_distance # Initialize the adjacency list for the graph graph = [[] for _ in range(n + 1)] for u, v in edges: graph[u].append(v) graph[v].append(u) # Perform BFS from node 1 (the water source) return bfs(1)