题目
题目链接
解题思路
- 解析输入数据，获取玩家坐标和所有 \(NPC\) 坐标
- 计算玩家到每个 \(NPC\) 的欧几里得距离
- 找到最短距离对应的 \(NPC\) 坐标
- 按要求格式输出结果
代码
#include <iostream>
#include <vector>
#include <string>
#include <sstream>
#include <cmath>
using namespace std;int main() {// 读取玩家坐标和NPC数量int playerX, playerY, n;char comma;cin >> playerX >> comma >> playerY >> comma >> n >> comma;// 读取NPC坐标vector<pair<int, int>> npcs;string input;cin >> input;stringstream ss(input);for(int i = 0; i < n; i++) {int x, y;ss >> x >> comma >> y;if(i < n-1) ss >> comma;npcs.push_back({x, y});}// 找最近的NPCdouble minDist = 1e9;int resX = 0, resY = 0;for(auto& npc : npcs) {double dist = sqrt(pow(playerX - npc.first, 2) + pow(playerY - npc.second, 2));if(dist < minDist) {minDist = dist;resX = npc.first;resY = npc.second;}}// 输出结果cout << "(" << resX << "," << resY << ")" << endl;return 0;
}
import java.util.*;public class Main {public static void main(String[] args) {Scanner sc = new Scanner(System.in);// 去除所有空格后再按逗号分割String[] input = sc.nextLine().replaceAll("\\s+", "").split(",");try {// 解析玩家坐标和NPC数量int playerX = Integer.parseInt(input[0]);int playerY = Integer.parseInt(input[1]);int n = Integer.parseInt(input[2]);// 找最近的NPCdouble minDist = Double.MAX_VALUE;int resX = 0, resY = 0;// 遍历所有NPCfor(int i = 0; i < n; i++) {int npcX = Integer.parseInt(input[3 + i*2]);int npcY = Integer.parseInt(input[4 + i*2]);// 确保坐标在地图范围内if(npcX >= 0 && npcX < 128 && npcY >= 0 && npcY < 128) {double dist = Math.sqrt(Math.pow(playerX - npcX, 2) + Math.pow(playerY - npcY, 2));if(dist < minDist) {minDist = dist;resX = npcX;resY = npcY;}}}// 输出结果System.out.printf("(%d,%d)\n", resX, resY);} catch(Exception e) {// 处理可能的数组越界或数字格式异常System.out.println("输入格式错误");}}
}
def solve():# 读取输入inputs = input().split(',')playerX = int(inputs[0])playerY = int(inputs[1])n = int(inputs[2])# 解析NPC坐标npcs = []for i in range(n):x = int(inputs[3 + i*2])y = int(inputs[4 + i*2])npcs.append((x, y))# 找最近的NPCmin_dist = float('inf')res_x = res_y = 0for x, y in npcs:dist = ((playerX - x) ** 2 + (playerY - y) ** 2) ** 0.5if dist < min_dist:min_dist = distres_x, res_y = x, y# 输出结果print(f"({res_x},{res_y})")if __name__ == "__main__":solve()
算法及复杂度
- 算法：遍历计算欧几里得距离
- 时间复杂度：\(\mathcal{O}(n)\)，\(n\) 为 \(NPC\) 数量
- 空间复杂度：\(\mathcal{O}(n)\)，存储 \(NPC\) 坐标
