Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)

Total Submission(s): 8065    Accepted Submission(s): 2959

Problem Description

Angel was caught by the MOLIGPY! He was put in prison by Moligpy. The prison is described as a N * M (N, M <= 200) matrix. There are WALLs, ROADs, and GUARDs in the prison.

Angel's friends want to save Angel. Their task is: approach Angel. We assume that "approach Angel" is to get to the position where Angel stays. When there's a guard in the grid, we must kill him (or her?) to move into the grid. We assume that we moving up, down, right, left takes us 1 unit time, and killing a guard takes 1 unit time, too. And we are strong enough to kill all the guards.

You have to calculate the minimal time to approach Angel. (We can move only UP, DOWN, LEFT and RIGHT, to the neighbor grid within bound, of course.)

 

 

Input

First line contains two integers stand for N and M.

Then N lines follows, every line has M characters. "." stands for road, "a" stands for Angel, and "r" stands for each of Angel's friend.

Process to the end of the file.

 

 

Output

For each test case, your program should output a single integer, standing for the minimal time needed. If such a number does no exist, you should output a line containing "Poor ANGEL has to stay in the prison all his life."

 

 

Sample Input

7 8 #.#####. #.a#..r. #..#x... ..#..#.# #...##.. .#...... ........

 

 

Sample Output

13

 

 

Author

CHEN, Xue

 

 

Source

 

 

Recommend

Eddy

 

//广搜的应用

#include <iostream>

#include <stdio.h>

#include <queue>

#include <string.h>

#define N 203

using namespace std;

char map[N][N];

int r[N][N];

int dir[4][2]={0,1,0,-1,1,0,-1,0};

int n,m;

struct node

{

    int x,y;

};

bool is_ok(node &t)

{

    if(t.x<1||t.x>n||t.y<1||t.y>m||map[t.x][t.y]=='#')

       return false;

    return true;

}

void bfs(int x,int y)

{

    queue<node> Q;

    node a,b;

    int i,l;

    a.x=x;a.y=y;

    Q.push(a);

    while(!Q.empty())

    {

        a=Q.front();

        Q.pop();

       for(i=0;i<4;i++)

       {

           b.x=a.x+dir[i][0];

           b.y=a.y+dir[i][1];

           if(!is_ok(b)||b.x==x&&b.y==y)

             continue;

             l=1;

            if(map[b.x][b.y]=='x')

                l=2;

             if(r[b.x][b.y]==0)

              {

                r[b.x][b.y]=r[a.x][a.y]+l;

                Q.push(b);

              }

              else

              {

                  if(r[b.x][b.y]>r[a.x][a.y]+l)//发现有更近的、就替换掉

                    {

                        r[b.x][b.y]=r[a.x][a.y]+l;

                        Q.push(b);

                    }

              }

       }

    }

}

int main()

{

    int i,j;

    int x,y;

    int a_x,a_y;

    while(scanf("%d%d",&n,&m)!=EOF)

    {

        getchar();

        for(i=1;i<=n;getchar(),i++)

          for(j=1;j<=m;j++)

            {

                scanf("%c",&map[i][j]);

                if(map[i][j]=='r')

                 x=i,y=j;

                if(map[i][j]=='a')

                 a_x=i,a_y=j;

               r[i][j]=0;

            }

       bfs(x,y);

       if(r[a_x][a_y]!=0)

         printf("%d\n",r[a_x][a_y]);

        else

         printf("Poor ANGEL has to stay in the prison all his life.\n");//开始忘记写这个了，又贡献了次//WA，唉、、自己都无语了

    }

    return 0;

}