諸々の原因により、提出がぎりぎり間に合わず。

solution

$2 \times H \times W \le 500000 = 5 \times 10^5$なので、knight/bishopの別と座標に関しての全探索が間に合う。$O(HW)$。

implementation

#include <iostream>
#include <vector>
#include <queue>
#include <tuple>
#include <functional>
#define repeat(i,n) for (int i = 0; (i) < (n); ++(i))
using namespace std;
const vector<int> knight_move_y { 1, 2, 2, 1, -1, -2, -2, -1 };
const vector<int> knight_move_x { 2, 1, -1, -2, -2, -1, 1, 2 };
const vector<int> bishop_move_y { 1, 1, -1, -1 };
const vector<int> bishop_move_x { 1, -1, -1, 1 };
const int inf = 1e9+7;
int main() {
    // input
    int h, w; cin >> h >> w;
    vector<string> s(h); repeat (y,h) cin >> s[y];
    // prepare
    auto on_field = [&](int y, int x) { return 0 <= y and y < h and 0 <= x and x < w; };
    int gy, gx, sy, sx;
    repeat (y,h) repeat (x,w) {
        switch (s[y][x]) {
            case 'S': sy = y; sx = x; break;
            case 'G': gy = y; gx = x; break;
        }
    }
    // bfs
    vector<vector<vector<int> > > dp(2, vector<vector<int> >(h, vector<int>(w, inf)));
    queue<tuple<bool,int,int> > que;
    que.push(make_tuple(false, gy, gx));
    dp[0][gy][gx] = 0;
    while (not que.empty()) {
        bool bishop; int y, x; tie(bishop, y, x) = que.front(); que.pop();
        vector<int> const & dy = bishop ? bishop_move_y : knight_move_y;
        vector<int> const & dx = bishop ? bishop_move_x : knight_move_x;
        int len = dy.size();
        repeat (i,len) {
            int ny = y + dy[i];
            int nx = x + dx[i];
            if (not on_field(ny, nx)) continue;
            bool nbishop = s[ny][nx] == 'R' ? not bishop : bishop;
            if (dp[nbishop][ny][nx] == inf) {
                dp[nbishop][ny][nx] = dp[bishop][y][x] + 1;
                que.push(make_tuple(nbishop, ny, nx));
            }
        }
    }
    // output
    int ans = min(dp[0][sy][sx], dp[1][sy][sx]);
    if (ans == inf) ans = -1;
    cout << ans << endl;
    return 0;
}