Rescue（逆向BFS）

Link：http://acm.hdu.edu.cn/showproblem.php?pid=1242

Problem：

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 16584    Accepted Submission(s): 6025

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

ZOJ Monthly, October 2003

 

code：

#include
#include
#include
using namespace std;
char map[201][201];
int vis[201][201],ex,ey,n,m;
int d[4][2]={{-1,0},{0,-1},{1,0},{0,1}};
struct node{
int x;
int y;
int step;
};
int bfs()
{
queueQ;
memset(vis,0,sizeof(vis));
int i,xi,yi;
node now,next;
now.step=0;
now.x=ex;
now.y=ey;
Q.push(now);
vis[ex][ey]=1;
while(!Q.empty())
{
now=Q.front();
Q.pop();
if(map[now.x][now.y]=='r')
return now.step;
for(i=0;i<4;i++)
{
xi=now.x+d[i][0];
yi=now.y+d[i][1];
if(xi>=0&&xi=0&&yi{
vis[xi][yi]=1;
next.x=xi;
next.y=yi;
if(map[xi][yi]=='x')
next.step=now.step+2;
else
next.step=now.step+1;
Q.push(next);
}
}
}
return -1;
}
int main()
{
int i,j,fg;
while(scanf("%d%d",&n,&m)!=EOF)
{
for(i=0;i{
scanf("%s",map[i]);
for(j=0;j{
 if(map[i][j]=='a')
{
ex=i;
ey=j;
}
}
}
fg=bfs();
if(fg==-1)
printf("Poor ANGEL has to stay in the prison all his life.\n");
else
printf("%d\n",fg);
}
return 0;
}

总结：此题注意可能有多个’r’，而’a’只有一个，从’a’开始搜，找到的第一个’r’即为所求