Codeforces Good Bye 2017 B (BF算法也就是枚舉法的一種)

題目：B. New Year and Buggy Bot
Bob programmed a robot to navigate through a 2d maze.
The maze has some obstacles. Empty cells are denoted by the character '.', where obstacles are denoted by '#'.
There is a single robot in the maze. It's start position is denoted with the character 'S'. This position has no obstacle in it. There is also a single exit in the maze. It's position is denoted with the character 'E'. This position has no obstacle in it.
The robot can only move up, left, right, or down.
When Bob programmed the robot, he wrote down a string of digits consisting of the digits 0 to 3, inclusive. He intended for each digit to correspond to a distinct direction, and the robot would follow the directions in order to reach the exit. Unfortunately, he forgot to actually assign the directions to digits.
The robot will choose some random mapping of digits to distinct directions. The robot will map distinct digits to distinct directions. The robot will then follow the instructions according to the given string in order and chosen mapping. If an instruction would lead the robot to go off the edge of the maze or hit an obstacle, the robot will crash and break down. If the robot reaches the exit at any point, then the robot will stop following any further instructions.
Bob is having trouble debugging his robot, so he would like to determine the number of mappings of digits to directions that would lead the robot to the exit.
Input
The first line of input will contain two integers n and m (2 ≤ n, m ≤ 50), denoting the dimensions of the maze.
The next n lines will contain exactly m characters each, denoting the maze.
Each character of the maze will be '.', '#', 'S', or 'E'.
There will be exactly one 'S' and exactly one 'E' in the maze.
The last line will contain a single string s (1 ≤ |s| ≤ 100) — the instructions given to the robot. Each character of s is a digit from 0 to 3.
Output
Print a single integer, the number of mappings of digits to directions that will lead the robot to the exit.

樣例：
Examples
Input
5 6
.....#
S....#
.#....
.#....
...E..
333300012
Output
1
Input
6 6
......
......
..SE..
......
......
......
01232123212302123021
Output
14
Input
5 3
...
.S.
###
.E.
...
3
Output
0

提示：For the first sample, the only valid mapping is , where D is down, L is left, U is up, R is right；

題意：有一個地圖，其中S表示機器人開始的位置，E表示機器人結束的位置，#表示障礙，‘.’表示空的地方，原本0123分別表示上下左右，可是如今不知道哪個數字表示上，哪個數表示下等。。。；
以後第一行輸入兩個數表示輸入地圖的行和列，而後輸入地圖，再以後輸入一串字符，即0123，。而後讓你判斷這段字符可否在全部不一樣表示狀況下能走出來的次數（這裏的不一樣表示狀況就是上下左右可能爲0123，也可能爲1230，也可能爲1320，等）；

思路：將全部上下左右被0123組合表示的狀況列出來而後用一個數組存下來，而後再輸入的字符串中的0123的字符表示其下標，其內部的0123分別表示上下左右；而後移動S，而後判斷可否順利到達E；

新技巧：就是在數的組合狀況不少的時候，可是是已知多少的時候，若須要將每一種組合都進行一遍的話，則就能夠把這些數提早存到數組中；（和前面寫的那些字符串的枚舉法一個道理）；

代碼：

#include<stdio.h>
int mu[24][4]={{0,1,2,3},{0,1,3,2},{0,2,3,1},{0,2,1,3},{0,3,2,1},{0,3,1,2},{1,0,2,3},{1,0,3,2},{1,2,0,3},{1,2,3,0},{1,3,0,2},{1,3,2,0},{2,0,1,3},{2,0,3,1},{2,1,3,0},{2,1,0,3},{2,3,0,1},{2,3,1,0},{3,0,1,2},{3,0,2,1},{3,1,2,0},{3,1,0,2},{3,2,1,0},{3,2,0,1}};
int main()
{
    char s[55][55],a[105];
    int n,m,i,j,b[2][2],p,q,num;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
        scanf("%s",s[i]);
    scanf("%s",a);
    for(i=0;i<n;i++)
        for(j=0;j<m;j++)
        {
            if(s[i][j]=='S')
            {b[0][0]=i;b[0][1]=j;}
            if(s[i][j]=='E')
            {b[1][0]=i;b[1][1]=j;}
        }
    num=0;
    for(i=0;i<24;i++)
    {
        p=b[0][0];q=b[0][1];
        for(j=0;a[j]!='\0';j++)
        {
            switch(mu[i][a[j]-'0'])
            {
                case 0:p--;break;
                case 1:q--;break;
                case 2:p++;break;
                case 3:q++;break;
            }
            if(s[p][q]=='#')
                break;
            else if(p<0||q<0||p>=n||q>=m)
                break;
            else if(s[p][q]=='E')
            {
                num++;break;
            }
        }
    }
    printf("%d\n",num);
    return 0;
}