题目分析

题目要求我们根据给定的传统藏宝图，计算宝藏的精确位置和从起点到宝藏的直线距离。藏宝图由一系列方向和步数组成，每个方向代表八个基本罗盘方向之一。我们需要将这些步数和方向转换为坐标变化，最终计算宝藏的坐标和直线距离。

解题思路

1. 解析输入：每个藏宝图由一个字符串表示，包含多个逗号分隔的步数和方向组合，以句号结束。例如，"3N,1E,1N,3E,2S,1W."。
2. 方向处理：将每个方向转换为对应的坐标变化。例如：
   • N (北) → y += step
   • NE (东北) → x += step / √2, y += step / √2
   • E (东) → x += step
   • SE (东南) → x += step / √2, y -= step / √2
   • S (南) → y -= step
   • SW (西南) → x -= step / √2, y -= step / √2
   • W (西) → x -= step
   • NW (西北) → x -= step / √2, y += step / √2
3. 坐标计算：遍历每个步数和方向，更新当前坐标 (x, y)。
4. 距离计算：使用欧几里得距离公式计算从原点 (0, 0) 到 (x, y) 的距离，即 sqrt(x^2 + y^2)。
5. 输出结果：按照指定格式输出坐标和距离，保留三位小数。

正确代码

#include <iostream>
#include <string>
#include <vector>
#include <cmath>
#include <iomanip>
#include <sstream>
using namespace std;

struct Point {
    double x, y;
    Point() : x(0), y(0) {}
    void move(int steps, const string& dir) {
        if (dir == "N") y += steps;
        else if (dir == "NE") { x += steps/sqrt(2); y += steps/sqrt(2); }
        else if (dir == "E") x += steps;
        else if (dir == "SE") { x += steps/sqrt(2); y -= steps/sqrt(2); }
        else if (dir == "S") y -= steps;
        else if (dir == "SW") { x -= steps/sqrt(2); y -= steps/sqrt(2); }
        else if (dir == "W") x -= steps;
        else if (dir == "NW") { x -= steps/sqrt(2); y += steps/sqrt(2); }
    }
};

vector<string> split_commands(const string& s) {
    vector<string> commands;
    size_t start = 0;
    size_t end = s.find(',');
    while (end != string::npos) {
        commands.push_back(s.substr(start, end - start));
        start = end + 1;
        end = s.find(',', start);
    }
    commands.push_back(s.substr(start, s.size() - start - 1)); // 去掉最后的点
    return commands;
}

void process_map(const string& map_desc, int map_num) {
    Point treasure;
    vector<string> commands = split_commands(map_desc);
    
    for (const string& cmd : commands) {
        size_t dir_pos = 0;
        while (dir_pos < cmd.size() && isdigit(cmd[dir_pos])) {
            dir_pos++;
        }
        int steps = stoi(cmd.substr(0, dir_pos));
        string dir = cmd.substr(dir_pos);
        treasure.move(steps, dir);
    }
    
    double distance = sqrt(treasure.x * treasure.x + treasure.y * treasure.y);
    
    cout << "Map #" << map_num << endl;
    cout << fixed << setprecision(3);
    cout << "The treasure is located at (" << treasure.x << "," << treasure.y << ")." << endl;
    cout << "The distance to the treasure is " << distance << "." << endl;
    cout << endl;
}

int main() {
    string line;
    int map_num = 1;
    
    while (getline(cin, line)) {
        if (line == "END") break;
        process_map(line, map_num);
        map_num++;
    }
    
    return 0;
}