傳送門：http://codeforces.com/gym/101002
題意：有一個n*m個畫板，起始位置爲(n,1)，有q條指令，每條指令表示向上下左右一個方向走cnt步，每走一步花費1秒，每走到一個格子就會將格子染黑，現在給出目標圖形，並且可以選擇一段區間[L,R]使得在這段時間內走過的格子會變白，問這段區間L最小爲多少，R最大爲多少？
題解：對於目標圖形，假設某個格子座標是(x,y)，走到這個格子的最後時間爲time：
當mp[x][y]=='#'時，那麼L>time
當mp[x][y]=='.'時，那麼R<=time
只要按時間從後往前，找到每個格子被走到的最後時間，用來更新[L,R]即可
#include<bits/stdc++.h>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define x first
#define y second
#define rep(i,a,b) for(int i=(a);i<(b);++i)
#define per(i,a,b) for(int i=(a)-1;i>=(b);--i)
#define fuck(x) cout<<'['<<#x<<' '<<(x)<<']'<<endl
#define clr(a,b) memset(a,b,sizeof(a))
#define cp(a,b) memcpy(a,b,sizeof(b))
#define eps 1e-10
#define pb push_back
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef vector<int> VI;
typedef pair<int, int> pii;
typedef unsigned int ui;
const int INF = 0x3f3f3f3f;
const ll INFLL = 0x3f3f3f3f3f3f3f3fLL;
const int mod = 175781251;
const int MX = 1e6 + 5;
char str[2 * MX];
int n, m;
set<int>sx[MX], sy[MX];
char get(int x, int y) {
return str[x * m + y];
}
int to[][2] = {1, 0, 0, 1, 0, -1, -1, 0};
int op(char ch) {
if(ch == 'd') return 0;
if(ch == 'r') return 1;
if(ch == 'l') return 2;
return 3;
}
void solve(set<int>& s, int now, int l, int r, ll last, ll& L, ll& R, int mark) {
set<int>::iterator it = s.lower_bound(min(l, r));
for(; it != s.end() && *it <= max(l, r);) {
char ch = mark ? get(*it, now) : get(now, *it);
if(mark) sx[*it].erase(now);
else sy[*it].erase(now);
if(ch == '#') L = max(L, last - abs(*it - l));
else R = min(R, last - abs(*it - l) - 1);
s.erase(it++);
}
}
int t[MX], c[MX];
int main() {
#ifdef local
freopen("in.txt", "r", stdin);
#endif // local
int q; cin >> n >> m >> q;
rep(i, 1, n + 1) scanf("%s", str + i * m + 1);
rep(i, 1, n + 1) rep(j, 1, m + 1) {
sx[i].insert(j);
sy[j].insert(i);
}
char s[10];
int cx = n, cy = 1;
ll cnt = 1;
rep(i, 0, q) {
scanf("%s%d", s, &c[i]);
t[i] = op(s[0]);
cx += to[t[i]][0] * c[i];
cy += to[t[i]][1] * c[i];
cnt += c[i];
}
ll L = 0, R = cnt;
per(i, q, 0) {
int tp = 3 - t[i];
int nx = cx + to[tp][0] * c[i], ny = cy + to[tp][1] * c[i];
if(tp == 0 || tp == 3) solve(sy[cy], cy, cx, nx, cnt, L, R, 1);
else solve(sx[cx], cx, cy, ny, cnt, L, R, 0);
cnt -= c[i];
cx = nx;
cy = ny;
}
int flag = 1;
for(int i = 1; i <= n; i++) {
for(int y : sx[i]) {
if(get(i, y) == '#') L = R + 1;
}
}
if(L <= R) printf("%lld %lld\n", L, R);
else puts("-1 -1");
return 0;
}