HDU 1242 Rescue DFS

Rescue

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 27740    Accepted Submission(s): 9811

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

 

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题意就是天使被抓住了，然后他的朋友去救他，每走一个格时间加1，如果中途遇到士兵，这需要增加一个时间去消灭它，这个题有两个坑点，一是不一定碰到士兵，二是朋友不一定一个，诶，我说我怎么一直wa，看了别人的博客才发现第二个问题的。

本来是用的bfs，然后不只是一个朋友，又改的DFS

ac代码

#include <stdio.h>
#include <string.h>
char map[205][205];
int n,m,dir[4][2]={1,0,0,1,0,-1,-1,0};
int a[2],min,len,vis[205][205];

void dfs(int x,int y,int len)
{
	int i;
	
	if(x<0||x>=n||y<0||y>=m)
		return ;
	if(len>min)
		return ;
	if(vis[x][y])
		return ;
	if(map[x][y]=='#')
		return ;
	if(map[x][y]=='r')
	{
		if(min>len)
			min=len;
		
		return ;
	}
	if(map[x][y]=='x')
		len++;
	vis[x][y]=1;
	for(i=0;i<4;i++)
	{
		dfs(x+dir[i][0],y+dir[i][1],len+1);
	}
	vis[x][y]=0;
}

int main()
{
	int i,j;
	
	while(~scanf("%d%d",&n,&m))
	{

		for(i=0;i<n;i++)
		{
			getchar();		
			scanf("%s",&map[i]);
			for(j=0;j<m;j++)
			{
				if(map[i][j]=='a')
				{
					a[0]=i;
					a[1]=j;
				}
			}
		}
		min=100000;
		len=0;
		memset(vis,0,sizeof(vis));
		dfs(a[0],a[1],len);
		
		if(min!=100000)
			printf("%d\n",min);
		else
			printf("Poor ANGEL has to stay in the prison all his life.\n");
	}
	
	return 0;
}