zoj 1649 Rescue

Rescue

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Time Limit: 2 Seconds      Memory Limit: 65536 KB

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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 

//裸的bfs
#include <stdio.h>
#include <queue>
#include <string.h>
#include <iostream>
#include <math.h>
#include <algorithm>
#include <vector>
#include <map>
#include <string>
#define maxn 210
#define MM 10000000
using namespace std;
int n,m;
int ax,ay;
int xx[]={1,-1,0,0};
int yy[]={0,0,1,-1};
int mintime[maxn][maxn];
struct point
{
    int x;  //方格的位置
    int y;
    int step; //走的步数
    int time; //花的时间
}; queue<point> Q;
point start,end;
char ww[maxn][maxn];  //走到每一个位置所花的最少时间。
void bfs(point p) // 从開始位置開始bfs
{
    Q.push(p);
    point t;
    while(!Q.empty())
    {
        t=Q.front();
        Q.pop();
        for(int i=0;i<4;i++)
        {     
             point tt;  //向第i个方向走一步后的位置
            int x=t.x+xx[i],y=t.y+yy[i];
            if(x<n && x>=0 && m>y && y>=0 && ww[x][y]!='#')
            {

                tt.x=x;
                tt.y=y;
                 tt.step=t.step+1;
                  tt.time=t.time+1;
                if(ww[x][y]=='x' || ww[x][y]=='X')  //杀死禁卫多花一个时间
                    tt.time++;

            }
            if(tt.time<mintime[x][y])
            {
                mintime[x][y]=tt.time;
                Q.push(tt);
            }
        }
    }
}
int main()
{
    while(~scanf("%d%d",&n,&m))
    {
       for(int i=0;i<n;i++)
       for(int j=0;j<m;j++)
       mintime[i][j]=MM;
        for(int i=0;i<n;i++)
            scanf("%s",ww[i]);
        for(int i=0;i<n;i++)
            for(int j=0;j<m;j++)
              if(ww[i][j]=='r')
            {
            start.x=i;
            start.y=j;
            start.step=0;
            start.time=0;
            mintime[i][j]=0;
            }
            else if(ww[i][j]=='a')
                ax=i,ay=j;
            bfs(start);
            if(mintime[ax][ay]<MM)
            printf("%d
",mintime[ax][ay]);
            else
                printf("Poor ANGEL has to stay in the prison all his life.
");
    }
    return 0;
}
/*
7 8
#.#####.
#.a#.xr.
#.x#xx..
.x#.xx##
#...##..
.#......
........
*/