Codeforces 589J Cleaner Robot(DFS)

Posted 2020-07-11

Cleaner Robot

Time Limit: 2000MS

Memory Limit: 524288KB

64bit IO Format: %I64d & %I64u

Submit Status

Description

Masha has recently bought a cleaner robot, it can clean a floor without anybody‘s assistance.

Schematically Masha‘s room is a rectangle, consisting of w × h square cells of size 1 × 1. Each cell of the room is either empty (represented by character ‘.‘), or occupied by furniture (represented by character ‘*‘).

A cleaner robot fully occupies one free cell. Also the robot has a current direction (one of four options), we will say that it looks in this direction.

The algorithm for the robot to move and clean the floor in the room is as follows:

1. clean the current cell which a cleaner robot is in;
2. if the side-adjacent cell in the direction where the robot is looking exists and is empty, move to it and go to step 1;
3. otherwise turn 90 degrees clockwise (to the right relative to its current direction) and move to step 2.

The cleaner robot will follow this algorithm until Masha switches it off.

You know the position of furniture in Masha‘s room, the initial position and the direction of the cleaner robot. Can you calculate the total area of the room that the robot will clean if it works infinitely?

Input

The first line of the input contains two integers, w and h(1 ≤ w, h ≤ 10) — the sizes of Masha‘s room.

Next w lines contain h characters each — the description of the room. If a cell of a room is empty, then the corresponding character equals ‘.‘. If a cell of a room is occupied by furniture, then the corresponding character equals ‘*‘. If a cell has the robot, then it is empty, and the corresponding character in the input equals ‘U‘, ‘R‘, ‘D‘ or ‘L‘, where the letter represents the direction of the cleaner robot. Letter ‘U‘ shows that the robot is looking up according to the scheme of the room, letter ‘R‘ means it is looking to the right, letter ‘D‘ means it is looking down and letter ‘L‘ means it is looking to the left.

It is guaranteed that in the given w lines letter ‘U‘, ‘R‘, ‘D‘ or ‘L‘ occurs exactly once. The cell where the robot initially stands is empty (doesn‘t have any furniture).

Output

In the first line of the output print a single integer — the total area of the room that the robot will clean if it works infinitely.

Sample Input

Input

2 3
U..
.*.

Output

4

Input

4 4
R...
.**.
.**.
....

Output

12

Input

3 4
***D
..*.
*...

Output

6

Hint

In the first sample the robot first tries to move upwards, it can‘t do it, so it turns right. Then it makes two steps to the right, meets a wall and turns downwards. It moves down, unfortunately tries moving left and locks itself moving from cell (1, 3) to cell (2, 3) and back. The cells visited by the robot are marked gray on the picture.

Source

2015-2016 ACM-ICPC, NEERC, Southern Subregional Contest (Online Mirror, ACM-ICPC Rules, Teams Preferred)

解法一:

//英文题，看错了题。。。

//深搜，记录第一次经过某点的方向，此后再经过此点时，判断和初次的方向是否一致，如果一致，说明陷入了重复，不必继续搜索
#include <iostream>
#include <cstdio>
#include <cstring>
#include <cctype>
#include <set>
#include <stack>
#include <vector>
#include <algorithm>
using namespace std;

char s[20][20]; 
int n, m, cnt;
int dx[] = {0, -1, 0, 1, 0}, dy[] = {0, 0, 1, 0, -1}; //控制方向
bool use[20][20]; //标记是否经过                               
int dirr[20][20]; //记录第一次经过某点时的方向
void dfs(int x, int y, int dir)
{
    if(use[x][y] == false) //第一次经过此点，加一，记录方向
    {
        cnt++;
        dirr[x][y] = dir;
    }
    use[x][y] = true;

    int nx, ny, dir1;
    for(int i = 0; i < 4; i++) //顺时针旋转，只要碰到点就进入下一层，然后不能再继续顺时针旋转
    {
        dir1 = (dir + i) % 4;
        if(dir1 == 0) dir1 = 4;
        nx = x + dx[dir1], ny = y + dy[dir1];
        if(nx >= 0 && nx < n && ny >= 0 && ny < m && s[nx][ny] == ‘.‘)
        {
            if(dirr[nx][ny] == dir1) break;
            dfs(nx, ny, dir1);
            break;
        }
    }
}

int main()
{
    int x, y;
    while(~ scanf("%d%d", &n, &m))
    {
        int dir;
        cnt = 0;
        memset(use, false, sizeof use);
        for(int i = 0; i < n; i++)
            for(int j = 0; j < m; j++)
            {
                scanf(" %c", &s[i][j]);
                if(s[i][j] == ‘U‘) x = i, y = j, dir = 1, s[i][j] = ‘.‘;
                else if(s[i][j] == ‘R‘) x = i, y = j, dir = 2, s[i][j] = ‘.‘;
                else if(s[i][j] == ‘D‘) x = i, y = j, dir = 3, s[i][j] = ‘.‘;
                else if(s[i][j] == ‘L‘) x = i, y = j, dir = 4, s[i][j] = ‘.‘;
            }
        dfs(x, y, dir);
        printf("%d\n", cnt);
    }

    return 0;
}

 解法二:

#include<stdio.h>
#include<string.h>

const int MAXN = 17;

 ///‘U‘, ‘R‘, ‘D‘ ‘L‘
int dir[4][2] = { {-1,0},{0,1},{1,0},{0,-1} };
int M, N;
char G[MAXN][MAXN];
bool v[MAXN][MAXN][4];

void DFS(int k, int x, int y, int &ans)
{
    if(G[x][y] == ‘.‘)
    {
        ans += 1;
        G[x][y] = ‘#‘;
    }

    for(int i=0; i<4; i++)
    {
        int nx = x+dir[(i+k)%4][0];
        int ny = y+dir[(i+k)%4][1];

        if(nx>=0&&nx<M && ny>=0&&ny<N && G[nx][ny]!=‘*‘)
        {
            if(v[nx][ny][(i+k)%4] == true)
                break;
            v[nx][ny][(i+k)%4] = true;
            DFS((i+k)%4, nx, ny, ans);
            break;
        }
    }
}

int main()
{
    while(scanf("%d%d", &M, &N) != EOF)
    {
        memset(v, 0, sizeof(v));

        int x, y, op;

        for(int i=0; i<M; i++)
        {
            scanf("%s", G[i]);
            for(int j=0; j<N; j++)
            {
                if(G[i][j] == ‘U‘)
                    op = 0, x=i, y=j;
                if(G[i][j] == ‘R‘)
                    op = 1, x=i, y=j;
                if(G[i][j] == ‘D‘)
                    op = 2, x=i, y=j;
                if(G[i][j] == ‘L‘)
                    op = 3, x=i, y=j;
            }
        }

        int ans = 1;
        DFS(op, x, y, ans);

        printf("%d\n", ans);
    }

    return 0;
}