【杭电1242】广搜+队列

本文介绍了一个关于救援任务的问题,需要从监狱中救出被囚禁的朋友Angel。监狱由墙体、道路和守卫构成,任务是计算从任意起点到Angel位置所需的最短时间,包括消灭守卫的时间。

摘要生成于 C知道 ,由 DeepSeek-R1 满血版支持, 前往体验 >

 Rescue
Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

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<iostream>
#include<cstring>
#include<queue>
using namespace std;
int vis[200][200],sx,sy,n,m,prove;
char map[200][200];
int dx[4]={1,0,-1,0};
int dy[4]={0,1,0,-1};
int judge(int x,int y)
{
	if(x>=n||x<0||y<0||y>=m)
	return 0;
	else
	return 1;
}
struct node
{
	int x,y; 
	int time;
}s,e;
queue<node>q;
void bfs()
{
	while(!q.empty())
	q.pop();
	memset(vis,0,sizeof(vis));
	s.time=0;s.x=sx;s.y=sy;
	q.push(s);//把所有r的位置存入队列 
	vis[s.x][s.y]=1;//标记为走过的 
	while(!q.empty())
	{
		s=q.front();
		q.pop();
		if(map[s.x][s.y]=='a')//找到Angle,输出时间 
		{
			cout<<s.time<<endl;
			prove=1;
		}
		if(map[s.x][s.y]=='x')//遇到警卫,先打败,再次进入队列 
		{
			map[s.x][s.y]='.';//变为道路 
			s.time+=1;
			q.push(s);
			continue;
		}
		for(int i=0;i<4;i++)
		{
			e.x=s.x+dx[i];
			e.y=s.y+dy[i];
			if(map[e.x][e.y]=='#'||judge(e.x,e.y)==0||vis[e.x][e.y]==1)
			continue;
			//else 
			e.time=s.time+1;
			q.push(e);
			vis[e.x][e.y]=1;
			
		}
	}
}
int main()
{
	while(cin>>n>>m)
	{
		for(int i=0;i<n;i++)
		for(int j=0;j<m;j++)
		{
			cin>>map[i][j];
			if(map[i][j]=='r')
			{
				sx=i;sy=j;//记录朋友的位置 ,不止一个 
			}
		}
		prove=0;
		bfs();
		if(prove==0)
		cout<<"Poor ANGEL has to stay in the prison all his life."<<endl;
	}
	return 0;
}


评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值