这题目还是比较神仙的
不知道如何老师不提示应该怎么想到
反正老师这节课就讲分块，。。。emm
首先m的数据范围很小，这给了我们希望
发现没有down这个指令(我打不出来qaq)，所以产生 (loop) 的唯一可能是连续的'><'
对行进行分块，首先预处理出每个位置能够走到哪里，时间复杂度 (O(NM))
修改只需要将该块的元素能够走到的位置全部改一遍就行了
查询，emm直接跑就行了
具体实现见代码
// By Hacheylight
#include <map>
#include <set>
#include <ctime>
#include <queue>
#include <stack>
#include <cmath>
#include <vector>
#include <bitset>
#include <cstdio>
#include <cctype>
#include <string>
#include <numeric>
#include <cstring>
#include <cassert>
#include <climits>
#include <cstdlib>
#include <iostream>
#include <algorithm>
#include <functional>
using namespace std ;
//#define int long long
#define rep(i, a, b) for (int i = (a); i <= (b); i++)
#define per(i, a, b) for (int i = (a); i >= (b); i--)
#define loop(it, v) for (auto it = v.begin(); it != v.end(); it++)
#define cont(i, x) for (int i = head[x]; i; i = e[i].nxt)
#define clr(a) memset(a, 0, sizeof(a))
#define ass(a, sum) memset(a, sum, sizeof(a))
#define lowbit(x) (x & -x)
#define all(x) x.begin(), x.end()
#define SC(t, x) static_cast <t> (x)
#define ub upper_bound
#define lb lower_bound
#define pqueue priority_queue
#define mp make_pair
#define pb push_back
#define pof pop_front
#define pob pop_back
#define fi first
#define se second
#define y1 y1_
#define Pi acos(-1.0)
#define iv inline void
#define enter cout << endl
#define siz(x) ((int)x.size())
#define file(x) freopen(x".in", "r", stdin),freopen(x".out", "w", stdout)
typedef long double ld ;
typedef long long ll ;
typedef unsigned long long ull ;
typedef pair <int, int> pii ;
typedef pair <ll, int> pli ;
typedef pair<int, pii> tri ;
typedef vector <int> vi ;
typedef vector <pii> vii ;
typedef vector <vi> vvi ;
typedef queue <int> qi ;
typedef queue <pii> qii ;
typedef set <int> si ;
typedef map <int, int> mii ;
typedef map <string, int> msi ;
const int N = 100010 ;
const int M = 15 ;
const int K = 320 ;
const int INF = 0x3f3f3f3f ;
const int iinf = 1 << 30 ;
const ll linf = 2e18 ;
const int mod = 1000000007 ;
const double eps = 1e-7 ;
void douout(double x){ printf("%lf
", x + 0.0000000001) ; }
template <class T> void print(T a) { cout << a << endl ; exit(0) ; }
template <class T> void chmin(T &a, T b) { if (a > b) a = b ; }
template <class T> void chmax(T &a, T b) { if (a < b) a = b ; }
void add(int &a, int b) { a = a + b < mod ? a + b : a + b - mod ; }
void sub(int &a, int b) { a = (a - b + mod) % mod ; }
void mul(int &a, int b) { a = (ll) a * b % mod ; }
int addv(int a, int b) { return (a += b) >= mod ? a -= mod : a ; }
int subv(int a, int b) { return (a -= b) < 0 ? a += mod : a ; }
int mulv(int a, int b) { return (ll) a * b % mod ; }
int read() {
int f = 1, x = 0 ;
char ch = getchar() ;
while (!isdigit(ch)) { if (ch == '-') f = -1 ; ch = getchar() ; }
while (isdigit(ch)) { x = x * 10 + ch -'0' ; ch = getchar() ; }
return x * f ;
}
int pw(int a, int b) {
int s = 1 ;
for (; b; b >>= 1, a = (ll) a * a % mod)
if (b & 1) s = (ll) s * a % mod ;
return s ;
}
int n, m, q, len, num ;
int l[K], r[K], bl[N] ;
int dp[N][M] ;
char s[N][M] ;
void build() {
len = sqrt(n), num = (n - 1) / len + 1 ;
rep(i, 1, n) bl[i] = (i - 1) / len + 1 ;
rep(i, 1, num) l[i] = (i - 1) * len + 1, r[i] = i * len ;
r[num] = n ;
}
int dfs(int x, int y) {
if (dp[x][y]) return dp[x][y] ;
if (s[x][y] == '^') {
int tx = x - 1, ty = y ;
if (tx < l[bl[x]]) dp[x][y] = tx * (m + 2) + ty ;
else dp[x][y] = dfs(tx, ty) ;
}
if (s[x][y] == '>') {
int tx = x, ty = y + 1 ;
if (ty == m + 1) dp[x][y] = tx * (m + 2) + ty ;
else if (s[tx][ty] == '<') dp[x][y] = dp[tx][ty] = -1 ;
else dp[x][y] = dfs(tx, ty) ;
}
if (s[x][y] == '<') {
int tx = x, ty = y - 1 ;
if (ty == 0) dp[x][y] = tx * (m + 2) + ty ;
else if (s[tx][ty] == '>') dp[x][y] = dp[tx][ty] = -1 ;
else dp[x][y] = dfs(tx, ty) ;
}
return dp[x][y] ;
}
void block(int id) {
rep(i, l[id], r[id])
rep(j, 1, m)
dfs(i, j) ;
}
void modify(int x, int y, char c) {
rep(i, l[bl[x]], r[bl[x]])
rep(j, 1, m)
dp[i][j] = 0 ;
s[x][y] = c ;
block(bl[x]) ;
}
int query(int x, int y) {
if (dp[x][y] == -1) return -1 ;
if (bl[x] == 1) return dp[x][y] ; // 第一行
int tx = dp[x][y] / (m + 2), ty = dp[x][y] % (m + 2) ;
if (ty == 0 || ty == m + 1) return dp[x][y] ;
return query(tx, ty) ;
}
signed main() {
scanf("%d%d%d", &n, &m, &q) ;
rep(i, 1, n) scanf("%s", s[i] + 1) ;
build() ;
rep(i, 1, num) block(i) ;
while (q--) {
int x, y ; string op, ch ; cin >> op ;
if (op[0] == 'A') {
scanf("%d%d", &x, &y) ;
int pos = query(x, y) ;
if (pos == -1) puts("-1 -1") ;
else printf("%d %d
", pos / (m + 2), pos % (m + 2)) ;
} else {
scanf("%d%d", &x, &y) ; cin >> ch ;
modify(x, y, ch[0]) ;
}
}
return 0 ;
}