## 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;
}