题解 | #坐标移动#

主要思路：按分号作为分隔点，获取所有合法的坐标移动序列，再按坐标顺序模拟一遍即可。

#include <iostream>
#include <algorithm>
#include <string>
#include <cctype>

using namespace std;

const int N = 6000;

string a[N];

int main()
{
    string s;
    cin >> s;
    
    //cnt 记录合法移动序列的个数，last 记录上一个分号的下一个位置
    int cnt = 0, last = 0;  
    for (int i = 0; i < s.size(); i ++)
    {
        if (s[i] != ';') continue;
        string tmp = s.substr(last, i - last);
        if (tmp[0] == 'W' || tmp[0] == 'A' || tmp[0] == 'S' || tmp[0] == 'D')
        {
            bool f = true;
            for (int i = 1; i < tmp.size(); i ++)
                if (!isdigit(tmp[i])) f = false;
            if (f) a[++ cnt] = tmp;
        }
        last = i + 1;
    }
    
    int x = 0, y = 0;
    for (int i = 1; i <= cnt; i ++)
    {
        char d = a[i][0];
        int num = stoi(a[i].substr(1, a[i].size() - 1));
        if (d == 'W') y += num;
        else if (d == 'S') y-= num;
        else if (d == 'A') x -= num;
        else x += num;
    }
    
    cout << x << ',' << y << endl;
    
    return 0;
}