Problem I

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

[email protected]

5 4

JA#X

JVBN

XOHD

DQEM

[email protected]

Sample Output

forth forth right right forth forth forth

right forth forth left forth forth right

Problem-setter: Juergen Werner, Albert Einstein University of ULM, Germany

题目大意：

给你一个矩阵，只能按照IEHOVA的顺序走，输出轨迹的方向。@为起始点，#为终止点。

解题思路:

DFS,注意是从下往上走，i是减一的。

代码：

#include<iostream>
#include<cstdio>
#include<string>

using namespace std;

const int maxN=10;
const char letter[8]={'@','I','E','H','O','V','A','#'};
const string str0[3]={"left","forth","right"};
int dirX[3]={-1,0,1},n,m;
int dirY[3]={0,-1,0};//i是减不是加。
string mymap[maxN];

bool cmp(int di,int dj){
    if(di>=0&&di<n&&dj>=0&&dj<m) return true;
    return false;
}

void dfs(int i,int j,int depth){
    for(int k=0;k<3;k++){
        int di=i+dirY[k],dj=j+dirX[k];
        if(cmp(di,dj)&&mymap[di][dj]==letter[depth]){
            cout<<str0[k];
            if(depth<=6) cout<<" ";
            dfs(di,dj,depth+1);
        }
    }
}

int main(){
    int t,pi,pj;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        for(int i=0;i<n;i++){
            cin>>mymap[i];
            for(int j=0;j<m;j++){
                if(mymap[i][j]==letter[0]) pi=i,pj=j;
            }
        }
        dfs(pi,pj,1);
        cout<<endl;
    }
    return 0;
}
#include<iostream>
#include<cstdio>
#include<string>

using namespace std;

const int maxN=10;
const char letter[8]={'@','I','E','H','O','V','A','#'};
const string str0[3]={"left","forth","right"};
int dirX[3]={-1,0,1},n,m;
int dirY[3]={0,-1,0};//i是减不是加。
string mymap[maxN];

bool cmp(int di,int dj){
    if(di>=0&&di<n&&dj>=0&&dj<m) return true;
    return false;
}

void dfs(int i,int j,int depth){
    for(int k=0;k<3;k++){
        int di=i+dirY[k],dj=j+dirX[k];
        if(cmp(di,dj)&&mymap[di][dj]==letter[depth]){
            cout<<str0[k];
            if(depth<=6) cout<<" ";
            dfs(di,dj,depth+1);
        }
    }
}

int main(){
    int t,pi,pj;
    scanf("%d",&t);
    while(t--){
        scanf("%d %d",&n,&m);
        for(int i=0;i<n;i++){
            cin>>mymap[i];
            for(int j=0;j<m;j++){
                if(mymap[i][j]==letter[0]) pi=i,pj=j;
            }
        }
        dfs(pi,pj,1);
        cout<<endl;
    }
    return 0;
}