Rescue

Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)

Total Submission(s) : 4   Accepted Submission(s) : 2

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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

Source

ZOJ Monthly, October 2003 想法：bfs+優先佇列 程式碼：

#include <stdio.h>
#include <string.h>
#include <queue>
using namespace std;
struct node
{
    int x,y,step;
    friend bool operator < (node n1,node n2)    //自定義優先順序。在優先佇列中，優先順序高的元素先出佇列。
    {
         return n1.step > n2.step;    //通過題意可知 step 小的優先順序高，需要先出隊。
    }
};
char map[205][205];
int vis[205][205];
int to[4][2]= {1,0,-1,0,0,1,0,-1};
int n,m,sx,sy,ex,ey,ans;
int check(int x,int y)
{
    if(x<0 || x>=n || y<0 || y>=m)
        return 1;
    if(vis[x][y] || map[x][y]=='#')
        return 1;
    return 0;
}
void bfs()
{
    int i;
    priority_queue<node> Q;
    node a,next;
    a.x = sx;
    a.y = sy;
    a.step = 1;
    vis[a.x][a.y]=1;
    Q.push(a);
    while(!Q.empty())
    {
        a = Q.top();
        Q.pop();
        if(map[a.x][a.y]=='a')
        {
            ans = a.step;
            return ;
        }
        for(i = 0; i<4; i++)
        {
            next = a;
            next.x+=to[i][0];
            next.y+=to[i][1];
            if(check(next.x,next.y))
                continue;
            if(map[next.x][next.y]=='.') next.step=a.step+1;
            if(map[next.x][next.y]=='x') next.step=a.step+2;
            vis[next.x][next.y] = 1;
            Q.push(next);
        }
    }
    ans = -1;
}
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        int i,j;
        for(i = 0; i<n; i++)
            scanf("%s",map[i]);
        for(i = 0; i<n; i++)
        {
            for(j = 0; j<m; j++)
            {
                if(map[i][j]=='r')
                {
                    sx = i;
                    sy = j;
                }
                else if(map[i][j]=='a')
                {
                    ex = i;
                    ey = j;
                }
            }
        }
        memset(vis,0,sizeof(vis));
        ans=0;
        bfs();
        if(ans!=-1)
         printf("%d\n",ans);
        else
         printf("Poor ANGEL has to stay in the prison all his life.\n");
    }

    return 0;
}