分析

• 将二维坐标转化为一维，然后判断是否存在两个点在同一个并查集中即可

#include <iostream>

using namespace std;

const int N = 40010;

int n, m;
int p[N];

// 将二维坐标转化为一维，要求从左上角为(0, 0)
int get(int x, int y) {
    return x * n + y;
}

int find(int x) {
    if (p[x] != x) p[x] = find(p[x]);
    return p[x];
}

int main() {

    cin >> n >> m;

    for (int i = 0; i < n * n; i++) p[i] = i;

    int res = 0;
    for (int i = 1; i <= m; i++) {
        int x, y;
        char d;
        cin >> x >> y >> d;
        x--, y--;
        int a = get(x, y);
        int b;
        if (d == 'D') b = get(x + 1, y);
        else b = get(x, y + 1);

        int pa = find(a), pb = find(b);
        if (pa == pb) {
            res = i;
            break;
        }
        p[pa] = pb;
    }

    if (!res) puts("draw");
    else cout << res << endl;

    return 0;
}