[ABC265C] Belt Conveyor

Problem Statement

We have a grid with $H$ horizontal rows and $W$ vertical columns. $(i, j)$ denotes the square at the $i$-th row from the top and $j$-th column from the left.

$(i,j)$ has a character $G_{i,j}$ written on it. $G_{i,j}$ is U, D, L, or R.

You are initially at $(1,1)$. You repeat the following operation until you cannot make a move.

Let $(i,j)$ be the square you are currently at.

If $G_{i,j}$ is U and $i \neq 1$, move to $(i-1,j)$.

If $G_{i,j}$ is D and $i \neq H$, move to $(i+1,j)$.

If $G_{i,j}$ is L and $j \neq 1$, move to $(i,j-1)$.

If $G_{i,j}$ is R and $j \neq W$, move to $(i,j+1)$.

Otherwise, you cannot make a move.

Print the square you end up at when you cannot make a move.

If you indefinitely repeat moving, print -1 instead.

Constraints

• $1 \leq H, W \leq 500$
• $G_{i,j}$ is U, D, L, or R.
• $H$ and $W$ are integers.

---

Input

Input is given from Standard Input in the following format:

$H$ $W$
$G_{1,1}G_{1,2}\dots G_{1,W}$
$G_{2,1}G_{2,2}\dots G_{2,W}$
$\vdots$
$G_{H,1}G_{H,2}\dots G_{H,W}$

Output

If you end up at $(i, j)$, print it in the following format:

$i$ $j$

If you indefinitely repeat moving, print -1.

---

Sample Input 1

2 3
RDU
LRU

Sample Output 1

1 3

You will move as $(1, 1) \to (1, 2) \to (2, 2) \to (2, 3) \to (1, 3)$, ending up here, so the answer is $(1, 3)$.

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Sample Input 2

2 3
RRD
ULL

Sample Output 2

-1

You will indefinitely repeat moving as $(1, 1) \to (1, 2) \to (1, 3) \to (2, 3) \to (2, 2) \to (2, 1) \to (1, 1) \to (1, 2) \to \dots$, so -1 should be printed in this case.

---

Sample Input 3

9 44
RRDDDDRRRDDDRRRRRRDDDRDDDDRDDRDDDDDDRRDRRRRR
RRRDLRDRDLLLLRDRRLLLDDRDLLLRDDDLLLDRRLLLLLDD
DRDLRLDRDLRDRLDRLRDDLDDLRDRLDRLDDRLRRLRRRDRR
DDLRRDLDDLDDRLDDLDRDDRDDDDRLRRLRDDRRRLDRDRDD
RDLRRDLRDLLLLRRDLRDRRDRRRDLRDDLLLLDDDLLLLRDR
RDLLLLLRDLRDRLDDLDDRDRRDRLDRRRLDDDLDDDRDDLDR
RDLRRDLDDLRDRLRDLDDDLDDRLDRDRDLDRDLDDLRRDLRR
RDLDRRLDRLLLLDRDRLLLRDDLLLLLRDRLLLRRRRLLLDDR
RRRRDRDDRRRDDRDDDRRRDRDRDRDRRRRRRDDDRDDDDRRR

从点 $(1,1)$ 开始走，走到不能走为止。如果走到一个到过的点，说明有环，可以不停走，判断即可。

#include<cstdio>
const int tx[]={-1,1,0,0},ty[]={0,0,-1,1},N=505;
int n,m,v[N][N],t[N][N],x,y,dx,dy;
char ch;
int can(int x,int y)
{
	return x>0&&x<=n&&y>0&&y<=m;
}
int main()
{
	scanf("%d%d",&n,&m);
	for(int i=1;i<=n;i++)
	{
		for(int j=1;j<=m;j++)
		{
			scanf(" %c",&ch);
			if(ch=='D')
				t[i][j]=1;
			else if(ch=='L')
				t[i][j]=2;
			else if(ch=='R')
				t[i][j]=3;
		}
	}
	x=y=v[1][1]=1;
	while(1)
	{
		dx=x+tx[t[x][y]],dy=y+ty[t[x][y]];
		if(!can(dx,dy))
		{
			printf("%d %d\n",x,y);
			return 0;
		}
		x=dx,y=dy;
		if(v[x][y])
		{
			printf("-1");
			return 0;
		}
		v[x][y]=1;
	}
}