FZU 2150 fire game (bfs)
Accept: 2133 Submit: 7494
Time Limit: 1000 mSec Memory Limit :
32768 KB
Problem Description
Fat brother and Maze are playing a kind of special (hentai) game on an N*M
board (N rows, M columns). At the beginning, each grid of this board is
consisting of grass or just empty and then they start to fire all the grass.
Firstly they choose two grids which are consisting of grass and set fire. As we
all know, the fire can spread among the grass. If the grid (x, y) is firing at
time t, the grid which is adjacent to this grid will fire at time t+1 which
refers to the grid (x+1, y), (x-1, y), (x, y+1), (x, y-1). This process ends
when no new grid get fire. If then all the grid which are consisting of grass is
get fired, Fat brother and Maze will stand in the middle of the grid and playing
a MORE special (hentai) game. (Maybe it’s the OOXX game which decrypted in the
last problem, who knows.)
You can assume that the grass in the board would never burn out and the empty
grid would never get fire.
Note that the two grids they choose can be the same.
Input
The first line of the date is an integer T, which is the number of the text
cases.
Then T cases follow, each case contains two integers N and M indicate the
size of the board. Then goes N line, each line with M character shows the board.
“#” Indicates the grass. You can assume that there is at least one grid which is
consisting of grass in the board.
1 <= T <=100, 1 <= n <=10, 1 <= m <=10
Output
For each case, output the case number first, if they can play the MORE
special (hentai) game (fire all the grass), output the minimal time they need to
wait after they set fire, otherwise just output -1. See the sample input and
output for more details.
Sample Input
3 3
.#.
###
.#.
3 3
.#.
#.#
.#.
3 3
...
#.#
...
3 3
###
..#
#.#
Sample Output
Case 2: -1
Case 3: 0
Case 4: 2
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define INF 0x3f3f3f using namespace std; struct node{
int a,b;
int step;
};
node que1[];
int start1=,endd1=;
node grass[];
int couGrass=;
char ch[][];
int directX[]={-,,,};
int directY[]={,,-,};
int t,n,m; int bfs(int first,int second,int cougra){
int nowCouGra=;
int nowstep=;
start1=,endd1=;
char vis[][];
for(int i=;i<n;i++){
for(int j=;j<m;j++){
vis[i][j]=ch[i][j];
}
}
node t1,t2;
t1.a=grass[first].a;
t1.b=grass[first].b;
t1.step=;
t2.a=grass[second].a;
t2.b=grass[second].b;
t2.step=;
que1[endd1++]=t1;
que1[endd1++]=t2;
vis[t1.a][t1.b]='.';
vis[t2.a][t2.b]='.';
nowCouGra++,nowCouGra++;
while(start1<endd1){
node now=que1[start1++];
for(int i=;i<;i++){
node next;
next.a=now.a+directX[i];
next.b=now.b+directY[i];
next.step=now.step+;
if(next.a>=&&next.a<n&&next.b>=&&next.b<m){
if(vis[next.a][next.b]=='#'){
vis[next.a][next.b]='.';
que1[endd1++]=next;
nowstep=max(nowstep,next.step);
nowCouGra++;
}
}
}
if(nowCouGra==cougra){
return nowstep;
}
}
return INF;
} int main()
{
scanf("%d",&t);
for(int ii=;ii<t;ii++){
couGrass=;
scanf("%d %d",&n,&m);
getchar();
for(int j=;j<n;j++){
for(int k=;k<m;k++){
scanf("%c",&ch[j][k]);
if(ch[j][k]=='#'){
grass[couGrass].a=j;
grass[couGrass++].b=k;
}
}
getchar();
}
if(couGrass<=){
printf("Case %d: 0\n",ii+);
continue;
}
int ans=INF;
for(int i=;i<couGrass;i++){
for(int j=i+;j<couGrass;j++){
ans=min(ans,bfs(i,j,couGrass));
}
}
if(ans==INF){
printf("Case %d: -1\n",ii+);
}else{
printf("Case %d: %d\n",ii+,ans);
} }
return ;
}
FZU 2150 fire game (bfs)的更多相关文章
- FZU 2150 Fire Game(BFS)
点我看题目 题意 :就是有两个熊孩子要把一个正方形上的草都给烧掉,他俩同时放火烧,烧第一块的时候是不花时间的,每一块着火的都可以在下一秒烧向上下左右四块#代表草地,.代表着不能烧的.问你最少花多少时间 ...
- FZU Problem 2150 Fire Game(bfs)
这个题真要好好说一下了,比赛的时候怎么过都过不了,压点总是出错(vis应该初始化为inf,但是我初始化成了-1....),wa了n次,后来想到完全可以避免这个问题,只要入队列的时候判断一下就行了. 由 ...
- FZU 2150 Fire Game(点火游戏)
FZU 2150 Fire Game(点火游戏) Time Limit: 1000 mSec Memory Limit : 32768 KB Problem Description - 题目描述 ...
- FZU 2150 Fire Game (暴力BFS)
[题目链接]click here~~ [题目大意]: 两个熊孩子要把一个正方形上的草都给烧掉,他俩同一时候放火烧.烧第一块的时候是不花时间的.每一块着火的都能够在下一秒烧向上下左右四块#代表草地,.代 ...
- 【FZU - 2150】Fire Game(bfs)
--> Fire Game 直接写中文了 Descriptions: 两个熊孩子在n*m的平地上放火玩,#表示草,两个熊孩子分别选一个#格子点火,火可以向上向下向左向右在有草的格子蔓延,点火的地 ...
- FZU 2150 Fire Game (高姿势bfs--两个起点)(路径不重叠:一个队列同时跑)
Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows ...
- FZU 2150 Fire Game (高姿势bfs--两个起点)
Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board (N rows ...
- foj 2150 Fire Game(bfs暴力)
Problem Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M ...
- Fire Game--FZU2150(bfs)
http://acm.fzu.edu.cn/problem.php?pid=2150 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=659 ...
随机推荐
- Updating and Publishing a NuGet Package - Plus making NuGet packages smarter and avoiding source edits with WebActivator
I wrote a post a few days ago called "Creating a NuGet Package in 7 easy steps - Plus using NuG ...
- Unity游戏开发图片纹理压缩方案
Unity3D引擎对纹理的处理是智能的:不论你放入的是PNG,PSD还是TGA,它们都会被自动转换成Unity自己的Texture2D格式. 在Texture2D的设置选项中,你可以针对不同的平台,设 ...
- Nginx配置,413 Request Entity Too Large错误解决
今天有同事找我,说图片上传之后,不知道去哪里了.分析了一下问题,找到原因之后做了处理,这里简要记录一下. 问题原因: 1.首先后台log并无错误信息: 2.捡查了一下浏览器,发现network中有报错 ...
- C# System.IO.StreamReader
实现一个 TextReader,使其以一种特定的编码从字节流中读取字符. using System; using System.IO; class Test { public static void ...
- servlet的xx方式传值中文乱码
protected void doGet(HttpServletRequest req, HttpServletResponse resp) throws ServletException, IOEx ...
- NOIP2012 普及组 寻宝
题目描述 Description 传说很遥远的藏宝楼顶层藏着诱人的宝藏.小明历尽千辛万苦终于找到传说中的这个藏宝楼,藏宝楼的门口竖着一个木板,上面写有几个大字:寻宝说明书.说明书的内容如下: 藏宝楼共 ...
- 【MySQL】MySQL查询数据库各表的行数
#倒序查询数据库[各表记录数] use information_schema; select table_name,table_rows from tables where TABLE_SCHEMA ...
- [Memcached]分布式缓存系统Memcached在Asp.net下的应用
Memcached 是一个高性能的分布式内存对象缓存系统,用于动态Web应用以减轻数据库负载.它通过在内存中缓存数据和对象来减少读取数据库的次数,从而提高动态.数据库驱动网站的速度.Memcached ...
- WPF Button 样式
WPF CheckBox 自定义样式 给Button设置ToolTip <Style TargetType="{x:Type Button}" x:Key="Def ...
- Python PIL 的image类和numpy array之间的互换
import cv2 import numpy as np from PIL import Image from PIL import ImageEnhance def getline(frame): ...