题解 | #坐标移动#

画出dfa（确定性有限自动机）即可

// https://www.nowcoder.com/practice/119bcca3befb405fbe58abe9c532eb29

#include <iostream>
using namespace std;
using ii = pair<int, int>;
using handle = void (*)(int);
const int maxn = 5 + 1e4;

handle accept[maxn];
int trans[maxn][128], tsiz;  // t[u][x] = v: u --x--> v
ii dxs[128];

string s;  // input
ii ans;

void accept5(int p) {
    // a1;
    ii dx = dxs[s[p - 2]];
    int length = s[p - 1] - '0';
    dx.first *= length, dx.second *= length;
    ans.first += dx.first, ans.second += dx.second;
    // cout << s.substr(p - 2, 2) << endl;
    // printf("%d,%d\n", ans.first, ans.second);
}

void accept6(int p) {
    // a11;
    ii dx = dxs[s[p - 3]];
    int length = 10 * (s[p - 2] - '0') + s[p - 1] - '0';
    dx.first *= length, dx.second *= length;
    ans.first += dx.first, ans.second += dx.second;
    // cout << s.substr(p - 3, 3) << endl;
    // printf("%d,%d\n", ans.first, ans.second);
}

void init() {
    dxs['W'] = {0, 1};
    dxs['S'] = {0, -1};
    dxs['A'] = {-1, 0};
    dxs['D'] = {1, 0};
    trans[0][';'] = 1;  // 0 是非法状态，即如果没有转移，那么会转移到这个节点
    trans[1]['W'] = 2, trans[1]['A'] = 2, trans[1]['S'] = 2, trans[1]['D'] = 2;
    trans[1][';'] = 1;
    for (int i = '0'; i <= '9'; ++i) trans[2][i] = 3;
    trans[2][';'] = 1;
    for (int i = '0'; i <= '9'; ++i) trans[3][i] = 4;
    trans[3][';'] = 5;  // a1
    trans[4][';'] = 6;  // a11
    trans[5]['W'] = 2, trans[5]['A'] = 2, trans[5]['S'] = 2, trans[5]['D'] = 2;
    accept[5] = accept5;  // 5 是接受状态
    trans[6]['W'] = 2, trans[6]['A'] = 2, trans[6]['S'] = 2, trans[6]['D'] = 2;
    accept[6] = accept6;  // 6 是接受状态
}

int main() {
    init();
    cin >> s;
    int u = 1;
    for (int i = 0; i < s.size(); ++i) {
        u = trans[u][s[i]];
        // cout << u << " ";
        if (accept[u] != nullptr) accept[u](i);
    }
    printf("%d,%d", ans.first, ans.second);
}