Ignatius and the Princess I


Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 12711    Accepted Submission(s): 4018
Special Judge


Problem Description


The Princess has been abducted by the BEelzebub feng5166, our hero Ignatius has to rescue our pretty Princess. Now he gets into feng5166's castle. The castle is a large labyrinth. To make the problem simply, we assume the labyrinth is a N*M two-dimensional array which left-top corner is (0,0) and right-bottom corner is (N-1,M-1). Ignatius enters at (0,0), and the door to feng5166's room is at (N-1,M-1), that is our target. There are some monsters in the castle, if Ignatius meet them, he has to kill them. Here is some rules:

1.Ignatius can only move in four directions(up, down, left, right), one step per second. A step is defined as follow: if current position is (x,y), after a step, Ignatius can only stand on (x-1,y), (x+1,y), (x,y-1) or (x,y+1).
2.The array is marked with some characters and numbers. We define them like this:
. : The place where Ignatius can walk on.
X : The place is a trap, Ignatius should not walk on it.
n : Here is a monster with n HP(1<=n<=9), if Ignatius walk on it, it takes him n seconds to kill the monster.

Your task is to give out the path which costs minimum seconds for Ignatius to reach target position. You may assume that the start position and the target position will never be a trap, and there will never be a monster at the start position.


 


Input


The input contains several test cases. Each test case starts with a line contains two numbers N and M(2<=N<=100,2<=M<=100) which indicate the size of the labyrinth. Then a N*M two-dimensional array follows, which describe the whole labyrinth. The input is terminated by the end of file. More details in the Sample Input.


 



Output


For each test case, you should output "God please help our poor hero." if Ignatius can't reach the target position, or you should output "It takes n seconds to reach the target position, let me show you the way."(n is the minimum seconds), and tell our hero the whole path. Output a line contains "FINISH" after each test case. If there are more than one path, any one is OK in this problem. More details in the Sample Output.


 



Sample Input 


5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX.
5 6
.XX.1.
..X.2.
2...X.
...XX.
XXXXX1
5 6
.XX...
..XX1.
2...X.
...XX.
XXXXX.


 



Sample Output 


It takes 13 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
FINISH
It takes 14 seconds to reach the target position, let me show you the way.
1s:(0,0)->(1,0)
2s:(1,0)->(1,1)
3s:(1,1)->(2,1)
4s:(2,1)->(2,2)
5s:(2,2)->(2,3)
6s:(2,3)->(1,3)
7s:(1,3)->(1,4)
8s:FIGHT AT (1,4)
9s:FIGHT AT (1,4)
10s:(1,4)->(1,5)
11s:(1,5)->(2,5)
12s:(2,5)->(3,5)
13s:(3,5)->(4,5)
14s:FIGHT AT (4,5)
FINISH
God please help our poor hero.
FINISH

 

/*


第一次做优先队列的BFS 


*/ 


#include<cstdio>
#include<cstring>
#include<queue>
#define max 102
using namespace std;
 
struct Node
{
    int x,y,time;
    friend bool operator < (Node a,Node b)//运用重载比较运算符定义算子 
    {
         return a.time>b.time;//将队列的时间从大到小排列，小的先出列 
    }
};

struct node//路径存储每一步的前一步 
{
    int x,y,time;
}path[max][max];

char map[max][max];
int n,m,dir[][2]={0,1,0,-1,1,0,-1,0};

bool judge(int x,int y)//前进一步的判断条件 
{
    return x>=0&&y>=0&&x<n&&y<m&&map[x][y]!='X';
}

bool bfs(int ×)
{
    Node p,next;
    p.x=p.y=p.time=0;
    next.time=0;
    priority_queue<Node> q;
    q.push(p);
    while(q.size())
    {
        p=q.top();
        q.pop();
        if(p.x==n-1&&p.y==m-1)
        {
            times=p.time; return true;
        }
        for(int i=0; i<4; i++)
        {
            next=p;//注意，很巧妙的处理，为下面++next.time做铺垫 
            next.x+=dir[i][0];
            next.y+=dir[i][1];
            if(!judge(next.x,next.y))   continue;
            ++next.time;//路径存储过程 
            path[next.x][next.y].x=p.x;
            path[next.x][next.y].y=p.y;
            path[next.x][next.y].time=0;
            if(map[next.x][next.y]!='.')
            {
                next.time+=map[next.x][next.y]-'0';
                path[next.x][next.y].time=map[next.x][next.y]-'0';
            }
            map[next.x][next.y]='X';//标记为已路过 
            q.push(next);
        }
    }
    return false;
}

void printPath(int times,int x,int y)//输出路径
{ 
    if(times==0) return;
    times-=path[x][y].time;
    printPath(times-1,path[x][y].x,path[x][y].y);
    printf("%ds:(%d,%d)->(%d,%d)\n",times++,path[x][y].x,path[x][y].y,x,y);
    while(path[x][y].time--)
    {
        printf("%ds:FIGHT AT (%d,%d)\n",times++,x,y);
    }
}
int main()
{
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        for(int i=0; i<n; i++)
        {
            scanf("%s",map[i]);
        }
        memset(path,0,sizeof(path));
        int times=0;
        if(bfs(times))
        {
            printf("It takes %d seconds to reach the target position, let me show you the way.\n",times);
            printPath(times,n-1,m-1);
        }
        else
        {
            printf("God please help our poor hero.\n");
        }
        printf("FINISH\n");
    }
    return 0;
}