Fire逃生
Description:
You are trapped in a building consisting of open spaces and walls. Some places are on fire and you have to run for the exit. Will you make it?
At each second, the fire will spread to all open spaces directly connected to the North, South, East or West side of it. Fortunately, walls will never catch fire and will keep the fire inside the building, so as soon as you are out of the building you will be safe. To run to any of the four open spaces adjacent to you takes you exactly one second. You cannot run through a wall or into an open space that is on fire or is just catching fire, but you can run out of an open space at the same moment it catches fire.
Given a map of the building, decide how fast you can exit the building.
Input:
On the first line one positive number: the number of test cases, at most 100. After that per test case:
one line with two space-separated integers w and h (1 <= w, h <= 1 000): the width and height of the map of the building, respectively.
h lines with w characters each: the map of the building, consisting of
– ‘.’: a room,
– ‘#’: a wall,
– ‘@’: your starting location,
– ‘*’: fire.
There will be exactly one ‘@’ in the map.
Output:
Per test case:one line with a single integer which is the minimal number of seconds that you need to exit the building or the string “IMPOSSIBLE” when this is not possible.
思路:用两个队列分别存放人和大火所在位置,然后从队列取出,对每个‘@’和‘*’所在的点进行四个方向的拓展,对于‘@’,遇到‘.’则将‘.’替换成‘@’,并推进队列,表示人走过的安全点;对于‘*’,遇到‘.’和‘@’都替换成‘*’,并推进队列,表示火蔓延到该点。程序停止的条件为人已安全走出建筑物或被大火蔓延消失鸟。略挫,供参考=o=
/*0.28sec 4236 KB 3179 Bytes*/
#include <iostream>
#include <cstring>
#include <queue>
using namespace std; #define MAX 1010
typedef pair<int,int> Location; char room[MAX][MAX];
bool visit[MAX][MAX]; int dirt[][] = { {-,}, {,}, {,}, {,-}};
int x,y,costTime;
queue<Location> fire;
queue<Location> man; bool safeOrNot(int w,int h)
{
bool safe = false;
costTime = ;
while()
{
int man_Size = man.size();
int fire_Size = fire.size();
for(int k = ; k < fire_Size; k++)
{
Location cur = fire.front();
fire.pop();
for(int p = ; p < ; p++)
{
x = cur.first + dirt[p][];
y = cur.second + dirt[p][];
if( x >= && x < h && y >= && y < w && !visit[x][y])
{
if(room[x][y] == '.')
{
room[x][y] = '*';
//cout <<"room["<< x <<"]["<<y<<"] = '.'"<<endl;
fire.push(make_pair(x,y));
visit[x][y] = true;
}
}
else if(visit[x][y] && room[x][y] == '@')
{
room[x][y] = '*';
fire.push(make_pair(x,y));
}
}
}
for(int i = ; i < man_Size; i++)
{
Location tem = man.front();
man.pop();
for(int j = ; j < ; j++)
{
x = tem.first + dirt[j][];
y = tem.second + dirt[j][];
if( x >= && x < h && y >= && y < w && !visit[x][y])
{
if(room[x][y] == '.')
{
room[x][y] = '@';
man.push(make_pair(x,y));
visit[x][y] = true;
}
else
continue;
}
else if(x < || x >= h || y< || y >= w)
{
safe = true;
return safe;
}
}
}
costTime ++;
if(man.empty())
return safe;
}
} int main()
{
int T,w,h;
cin >> T;
while( T-- )
{
cin >> w >> h;
memset(room, , sizeof(room));
memset(visit,false,sizeof(visit));
while(!fire.empty())
{
fire.pop();
}
while(!man.empty())
{
man.pop();
}
for(int i = ; i < h; i++)
{
for(int j = ; j < w; j++)
{
cin >> room[i][j];
if(room[i][j] == '.')
visit[i][j] = false;
else
visit[i][j] = true;
if(room[i][j] == '@')
man.push(make_pair(i,j));
if(room[i][j] == '*')
fire.push(make_pair(i,j));
}
}
if(safeOrNot(w,h) )
cout << costTime << endl;
else
cout << "IMPOSSIBLE" << endl;
}
//system("pause");
return ;
}
Fire逃生的更多相关文章
- UVa 11624 Fire!(着火了!)
UVa 11624 - Fire!(着火了!) Time limit: 1.000 seconds Description - 题目描述 Joe works in a maze. Unfortunat ...
- CSUOJ 2031 Barareh on Fire
Description The Barareh village is on fire due to the attack of the virtual enemy. Several places ar ...
- CSU - 2031 Barareh on Fire (两层bfs)
传送门: http://acm.csu.edu.cn/csuoj/problemset/problem?pid=2031 Description The Barareh village is on f ...
- UVA11624 Fire! —— BFS
题目链接:https://vjudge.net/problem/UVA-11624 题解: 坑点:“portions of the maze havecaught on fire”, 表明了起火点不唯 ...
- 关于SequeezeNet中的Fire Module
在论文<SQUEEZENET: ALEXNET-LEVEL ACCURACY WITH 50X FEWER PARAMETERS AND <0.5MB MODEL SIZE>中,作者 ...
- FZU 2150 Fire Game
Fire Game Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%I64d & %I64u Submit St ...
- HDU 4857 逃生 (反向拓扑排序 & 容器实现)
题目链接:http://acm.hdu.edu.cn/showproblem.php?pid=4857 逃生 Time Limit: 2000/1000 MS (Java/Others) Mem ...
- Fire
Fire 分析: 首先,明确题意:b1,b2,--,bn 交换为b2,--,bn,b1,但这并不是意味着只能从b1开始交换,(这点从样例中可以看出),并且也不意味着交换的必须是连续的一串,可以是几个单 ...
- Android 轻量级输入校验库:Fire Eye
Fire Eye是一款轻量级简单易用的Android校验库. FireEye 2.0 在 1.0 的基础上,全部重写了代码,并优化了架构,性能上和逻辑上都大大提升.只需要几行代码,即可验证用户输入,并 ...
随机推荐
- 操作系统cmd算法
实验一 命令解释程序的编写(两周内) 一.目的和要求 1. 实验目的 (1)掌握命令解释程序的原理: (2)*掌握简单的DOS调用方法: (3)掌握C语言编程初步. 2.实验要求 编写类似于DOS, ...
- Windows安装配置php+memcached的方法
Windows下Memcached的安装配置方法 1.将第一个包解压放某个盘下面,比如在c:\memcached. 2.在终端(也即cmd命令界面)下输入 'c:\memcached\memcache ...
- JS图片加载失败显示默认图片
代码如下: <div id='photo<%# Container.DataItemIndex+1%>' style="position: absolute; displa ...
- ZooKeeper启动过程2:FastLeaderElection
前一篇文章中说到,启动ZooKeeper集群时,需要分别启动集群中的各个节点,各节点以QuorumPeer的形式启动,最后到达startLeaderElection和lookForLeader. 先说 ...
- vi notes
x = wqqq!, quit without save. movej,h,k,l^ or 0: start of line$: end of line:0, start of file:$, end ...
- 2013年8月份第4周51Aspx源码发布详情
迷你桌面闹钟源码 2013-8-27 [VS2010]功能介绍:实现了定时闹钟的功能,可以设置闹钟最前端显示.感兴趣的可以下载学习. BR个人博客系统(课程设计)源码 2013-8-27 [VS2 ...
- Thread启动方式一(Thread.start):源码分析
package day11; class TestDemo extends Thread{ int count = 0; /*public void add(){ while(count<100 ...
- 去掉NavigationBar底部的黑线
UINavigationBar *navigationBar = self.navigationController.navigationBar; [navigationBar setBackgr ...
- (spring-第2回【IoC基础篇】)Spring的Schema,基于XML的配置
要深入了解Spring机制,首先需要知道Spring是怎样在IoC容器中装配Bean的.而了解这一点的前提是,要搞清楚Spring基于Schema的Xml配置方案. 在深入了解之前,必须要先明白几个标 ...
- S50非接触式IC卡性能简介(M1)
一.主要指标 分为16个扇区,每个扇区为4块,每块16个字节,以块为存取单位: 每个扇区有独立的一组密码及访问控制: 每张卡有唯一序列号,为32位: 具有防冲突机制,支持多卡操作: 无电源,自带天线, ...