ZOJ - 1649 Rescue（dfs）

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

题意描述：先给出两个数字n和m然后再给出一个n*m的矩阵，a代表的是天使的位置，r代表的是朋友的位置，x代表士兵的位置，#代表墙的位置。r每走一步需要1s，r能杀死士兵且每杀死一个士兵也需要1s。问r能否到a，如果能输出所需要的最短时间。

剪枝的操作

if((step<book[tx][ty]||book[tx][ty]==0)&&e[tx][ty]!='#')
		{
			book[tx][ty]=step;
			if(e[tx][ty]=='x')
				dfs(tx,ty,step+2);
			else
				dfs(tx,ty,step+1);
		}

解题方法：在输入n*m的矩阵时记录r和a的位置，然后用book数组去存入r每次所走到这个位置时的时间step，判断这个点是否走过，如果走过就去和以前走到这个点的时间比较如果比以前到这里的时间短就去更新这个时间。

#include<stdio.h>
#include<string.h>

char e[210][210];
int book[210][210],n,m,flag=0,inf=999999;
int min=inf;
int a,b;

void dfs(int x,int y,int step)
{
	int k;
	int next[4][2]={1,0, -1,0, 0,1, 0,-1};
	if(min<step)
		return ;
	if(x==a&&y==b)
	{
		flag=1;
		if(min>step)
			min=step;
		return ;
	}

	int tx,ty;
	for(k=0;k<4;k++)
	{
		tx=x+next[k][0];
		ty=y+next[k][1];
		
		if(tx<0||ty<0||tx>=n||ty>=m)
			continue;
		if((step<book[tx][ty]||book[tx][ty]==0)&&e[tx][ty]!='#')
		{
			book[tx][ty]=step;
			if(e[tx][ty]=='x')
				dfs(tx,ty,step+2);
			else
				dfs(tx,ty,step+1);
		}
	}
}

int main(void)
{
	int x,y;
	while(~scanf("%d%d",&n,&m))
	{
		flag=0;min=inf;
		memset(book,0,sizeof(book));
		memset(e,0,sizeof(e));
		for(int i=0;i<n;i++)
			for(int j=0;j<m;j++)
			{
				scanf(" %c",&e[i][j]);
				if(e[i][j]=='r')
				{
					x=i;
					y=j;
				}
				if(e[i][j]=='a')
				{
					a=i;b=j;
				}
			}
		book[x][y]=0;
		dfs(x,y,0);
		if(flag==0)
			printf("Poor ANGEL has to stay in the prison all his life.\n");
		else
			printf("%d\n",min);
	}

	return 0;
 }