hdoj---Rescue

Rescue

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

Total Submission(s) : 60   Accepted Submission(s) : 22

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

 
 
 
 
 

#include<stdio.h>
#include<string.h>
#include<queue>
#include<algorithm>
#define N 210
using namespace std;
char a[N][N];
int b[N][N];
int n,m,ax,ay;
int dx[4]={0,1,-1,0};
int dy[4]={1,0,0,-1};
struct zz
{
	int x,y,ans;
	friend bool operator<(zz x,zz y)
	{
		return x.ans>y.ans;/*按照时间从小到大排序*/
	}
};
void bfs(int x,int y)
{
	memset(b,0,sizeof(b));
	priority_queue<zz>q;
	zz f1,f2;
	f1.x=x;f1.y=y;
	f1.ans=0;/*第一部一定要让ans=0*/
	q.push(f1);
	b[x][y]=1;
	while(!q.empty())
	{
		f1=q.top() ;
		q.pop();
		if(a[f1.x][f1.y]=='r')
		{
			printf("%d\n",f1.ans);
			return ;
		}
		for(int i=0;i<4;i++)
		{
			f2.x=f1.x+dx[i];
			f2.y=f1.y+dy[i];
			if(f2.x>0&&f2.x<=n&&f2.y>0&&f2.y<=m&&!b[f2.x][f2.y]&&a[f2.x][f2.y]!='#')
			{/*没有越界不是墙壁没有被用过的点执行下一步操作*/
				b[f2.x][f2.y]=1;
				if(a[f2.x][f2.y]=='x')
					f2.ans=f1.ans+2;/*遇到x时间加1*/
				else
					f2.ans=f1.ans+1;
					q.push(f2);
			}
		}
	}
	printf("Poor ANGEL has to stay in the prison all his life.\n");/*当遍历一遍之后还未找到人r*/
}
int main(){
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=1;i<=n;i++)
			scanf("%s",a[i]+1);
		for(i=1;i<=n;i++)
		{
			for(j=1;j<=m;j++)
			{
				if(a[i][j]=='a')
				{
					ax=i;ay=j;/*找到a的坐标*/
				}
			}
		}
		bfs(ax,ay);
	}
	return 0;
}