UVa, 10452 Marcus

Marcus, help!

Input: standard input
Output: standard output

Time Limit: 2 Seconds

"First, the breath of God.

Only the penitent man will pass.

Second, the Word of God,

Only in the footsteps of God will he proceed.

Third, the Path of God,

Only in the leap from the lion's head will he prove his worth."

(taken from the movie "Indiana Jones and the Last Crusade", 1989)

 

To get to the grail, Indiana Jones needs to pass three challenges. He successfully masters the first one and steps up to the second. A cobblestone path lies before him, each cobble is engraved with a letter. This is the second challenge, the Word of God, the Name of God. Only the cobbles having one of the letters "IEHOVA" engraved can be stepped on safely, the ones having a different letter will break apart and leave a hole.

 

Unfortunately, he stumbles and falls into the dust, and the dust comes into his eyes, making it impossible for him to see. So he calls for Marcus Brody and asks Marcus to tell him in which

direction to go to safely reach the other side of the cobblestone path. Because of the dust in his eyes, Indy only can step "forth" to the stone right in front of him or do a side-step to the stone on the "left" or the "right" side of him. These ("forth", "left", "right") are also the commands Marcus gives to him.

 

Input

The first line of the input contains a single number indicating the number of test cases that follow.

Each test case starts with a line containing two numbers m and n (2 <= m, n <= 8), the length m and the width n of the cobblestone path. Then follow m lines, each containing n characters ('A' to 'Z', '@', '#'), the engravement of the respective cobblestone. Indy's starting position is marked with the '@' character in the last line, the destination with the character '#' in the first line of the cobblestone path.

 

Each of the letters in "IEHOVA" and the characters '@' and '#' appear exactly once in each test case. There will always be exactly one path from Indy's starting position via the stones with the letters "IEHOVA" engraved on (in that order) to the destination. There will be no other way to safely reach the destination.

 

Output

For each test case, output a line with the commands Marcus gives to Indy so that Indy safely reaches the other side. Separate two commands by one space character.

 

Sample Input

2

6 5

PST#T

BTJAS

TYCVM

YEHOF

XIBKU

N@RJB

5 4

JA#X

JVBN

XOHD

DQEM

T@IY

 

Sample Output

forth forth right right forth forth forth

right forth forth left forth forth right

dfs的经典运用，题目要求按照"IEHOVA"的顺序走出迷宫，so，我只要定义一个字符串“"IEHOVA"”，用dfs一直往下走，直到走完字符串，输入和输出单独写的，用outland判断输入对不对，做题的时候遇到的问题，向左向右走没分清楚，还有就是输出的遍历所有的数组，后来发现，定义dfs的时候没有限定条件，当运行了几次之后不能进入dfs里了，之后就ac了。。。

 #include<iostream>

#include<vector>
#include<string>
#include<cstring>
#include<cstdio>


using namespace std;
/////////////////
const int dx[8] = {-1, -1, -1,  0, 0,  1, 1, 1};
const int dy[8] = {-1,  0,  1, -1, 1, -1, 0, 1};
const char a[8]={'@','I','E','H','O','V','A','#'};
//                0  1   2   3    4   5  6    7
class Marcus{
private:
    void dfs(int x,int y);
    bool visit[1000][1000];
    //vector<string>land;
    char mapl[1000][1000];
    int row,column;
    int x,y;
    int c;
public:
    void initial();
    void readcase();
    void computing();
    void outresult();
    void outland();
};


void Marcus::initial(){
    memset(mapl,0,sizeof(mapl));
    memset(visit,0,sizeof(visit));
    c=0;
}


void Marcus::readcase(){
    cin>>row>>column;
    //cout<<a[0]<<a[7];
    //string ch;
    for(int i=1;i<=row;i++){
        for(int j=1;j<=column;j++){
                cin>>mapl[i][j];
        //land.push_back(ch);
        }
    }
}
void Marcus::computing(){
    for(int i=1;i<=row;i++){
        for(int j=1;j<=column;j++){
            if(mapl[i][j]==a[0]){
                x=i;y=j;
                //cout<<x<<y<<endl;
                mapl[i][j]=true ;
                c++;
                //cout<<c;
            }
        }
    }
    dfs(x,y);




}
void Marcus::dfs(int x, int y){
    if( mapl[x][y+1] == a[c] && !visit[x][y+1]&&c<8){
            printf("right");
            if(c<7)printf(" ");
            visit[x][y+1]=true;
            c++;
            dfs(x,y+1);


    }
    if( mapl[x][y-1] == a[c] && !visit[x][y-1]&&c<8){
            printf("left");
            if(c<7)printf(" ");
            visit[x][y-1]=true;
            c++;
            dfs(x,y-1);


    }
    if( mapl[x-1][y] == a[c] && !visit[x-1][y]&&c<8 ){
            printf("forth");
            if(c<7)printf(" ");
            visit[x-1][y]=true;
            c++;
            dfs(x-1,y);


    }
    //if( mapl[x][y-1] == a[c] && !visit[x][y-1] ) c++;dfs(x,y-1);
}


void Marcus::outland(){
    for(int i=1;i<=row;i++){
        for(int j=1;j<=column;j++){
                cout<<mapl[i][j];
        }
        // cout<<land[i]<<endl;
        cout<<endl;
    }
}
////////////////////////////////////////////
int main(){
    int n;
    Marcus ma;
    cin>>n;
    while(n--){
        ma.initial();
        ma.readcase();
        ma.computing();
        //ma.outresult();
        //ma.outland();
        cout<<endl;
    }
    return 0;
}