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 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+ which refers to the grid (x+, y), (x-, y), (x, y+), (x, y-). 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. <= T <=, <= n <=, <= m <=
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 -. See the sample input and output for more details.
Sample Input
.#.
###
.#. .#.
#.#
.#. ...
#.#
... ###
..#
#.#
Sample Output
Case :
Case : -
Case :
Case :
暴力枚举两点,bfs统计,最后求最小值。
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<math.h>
#include<algorithm>
#include<queue>
#include<set>
#include<bitset>
#include<map>
#include<vector>
#include<stdlib.h>
#include <stack>
using namespace std;
#define PI acos(-1.0)
#define max(a,b) (a) > (b) ? (a) : (b)
#define min(a,b) (a) < (b) ? (a) : (b)
#define ll long long
#define eps 1e-10
#define MOD 1000000007
#define N 16
#define inf 1<<26
int n,m;
char mp[N][N];
int vis[N][N];
int dirx[]={,,-,};
int diry[]={-,,,};
struct Node{
int x,y;
int t;
}node[N*N]; int bfs(Node a,Node b){
queue<Node>q;
q.push(a);
q.push(b);
vis[a.x][a.y]=;
vis[b.x][b.y]=;
int T=;
while(!q.empty()){
Node tmp=q.front();
q.pop();
Node cnt;
T=max(T,tmp.t);
for(int i=;i<;i++){
cnt.x=tmp.x+dirx[i];
cnt.y=tmp.y+diry[i];
if(cnt.x< || cnt.x>=n || cnt.y< || cnt.y>=m) continue;
if(vis[cnt.x][cnt.y]) continue;
if(mp[cnt.x][cnt.y]=='.') continue;
vis[cnt.x][cnt.y]=;
cnt.t=tmp.t+;
q.push(cnt);
}
}
return T;
} int main()
{
int t,ac=;
scanf("%d",&t);
while(t--){
int num=;
scanf("%d%d",&n,&m);
for(int i=;i<n;i++){
scanf("%s",mp[i]);
for(int j=;j<m;j++){
if(mp[i][j]=='#'){
node[num].x=i;
node[num].y=j;
node[num++].t=;
}
}
} printf("Case %d: ",++ac);
if(num<=){
printf("0\n");
continue;
} int ans=inf;
for(int i=;i<num;i++){
for(int j=i;j<num;j++){
int flag=;
memset(vis,,sizeof(vis));
int w=bfs(node[i],node[j]);
//printf("===%d\n",w);
for(int k=;k<n;k++){
for(int l=;l<m;l++){
if(mp[k][l]=='#' && vis[k][l]==){
flag=;
break;
}
}
if(flag==){
break;
}
}
if(flag){
ans=min(ans,w);
}
}
} if(ans!=inf){
printf("%d\n",ans);
}
else{
printf("-1\n");
}
}
return ;
}
foj 2150 Fire Game(bfs暴力)的更多相关文章
- 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)
题目链接:http://acm.fzu.edu.cn/problem.php?pid=2150 Problem Description Fat brother and Maze are playing ...
- FZU 2150 Fire Game (bfs+dfs)
Problem Description Fat brother and Maze are playing a kind of special (hentai) game on an N*M board ...
- FZU - 2150 Fire Game bfs+双起点枚举
题意,10*10的地图,有若干块草地“#”,草地可以点燃,并在一秒后点燃相邻的草地.有墙壁‘·‘阻挡.初始可以从任意两点点火.问烧完最短的时间.若烧不完输出-1. 题解:由于100的数据量,直接暴力. ...
- fzu 2150 Fire Game 【身手BFS】
称号:fzupid=2150"> 2150 Fire Game :给出一个m*n的图,'#'表示草坪,' . '表示空地,然后能够选择在随意的两个草坪格子点火.火每 1 s会向周围四个 ...
- FZU 2150 fire game (bfs)
Problem 2150 Fire Game Accept: 2133 Submit: 7494Time Limit: 1000 mSec Memory Limit : 32768 KB ...
- FZU Problem 2150 Fire Game
Problem 2150 Fire Game Accept: 145 Submit: 542 Time Limit: 1000 mSec Memory Limit : 32768 KB P ...
- 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 ...
- FZOJ Problem 2150 Fire Game
...
随机推荐
- 【javascript 对日期的扩展 Format\addDays】
// 对Date的扩展,将 Date 转化为指定格式的String // 月(M).日(d).小时(H).分(m).秒(s).季度(q) 可以用 1-2 个占位符, // 年(y)可以用 1-4 个占 ...
- C#调用C++DLL传递结构体数组的终极解决方案
在项目开发时,要调用C++封装的DLL,普通的类型C#上一般都对应,只要用DllImport传入从DLL中引入函数就可以了.但是当传递的是结构体.结构体数组或者结构体指针的时候,就会发现C#上没有类型 ...
- (转载博文)MFC 窗口句柄获取
句柄获取方法(获取该窗口的句柄后,即可向该窗口类类发送消息.处理程序):0.获取所在类窗口的句柄: this->m_hwnd 1.主窗口的句柄: 无论在主窗口类内,还是子窗口类内,获取主窗口句柄 ...
- ID3决策树算法原理及C++实现(其中代码转自别人的博客)
分类是数据挖掘中十分重要的组成部分.分类作为一种无监督学习方式被广泛的使用. 之前关于"数据挖掘中十大经典算法"中,基于ID3核心思想的分类算法C4.5榜上有名.所以不难看出ID3 ...
- CGFW时装发布及活动整体一览表
CGFW时装发布及活动整体一览表 CGFW时装发布及活动整体一览表
- phpadmin
一晚上都在调试数据库,都要疯了,整理如下: 0.Apache服务器的443端口与VMware的冲突,所以要更改配置文件.设为440就可以(这个随意). 1.因为要远程访问,默认密码为空,所以首先给ro ...
- 步步学LINQ to SQL:使用LINQ检索数据【转】
[IT168 专稿]该系列教程描述了如何采用手动的方式映射你的对象类到数据表(而不是使用象SqlMetal这样的自动化工具)以便能够支持数据表之间的M:M关系和使用实体类的数据绑定.即使你选择使用了自 ...
- Leetcode:best_time_to_buy_and_sell_stock_II题解
一.题目 如果你有一个数组,它的第i个元素是一个股票在一天的价格. 设计一个算法,找出最大的利润. 二.分析 假设当前值高于买入值,那么就卖出,同一时候买入今天的股票,并获利.假设当前值低于买入值,那 ...
- PropertyGrid—为复杂属性提供下拉式编辑框和弹出式编辑框
零.引言 PropertyGrid中我们经常看到一些下拉式的编辑方式(Color属性)和弹出式编辑框(字体),这些都是为一些复杂的属性提供的编辑方式,本文主要说明如何实现这样的编辑方式. 一.为属性提 ...
- 30个你不可不知的CSS选择器
30个你不可不知的CSS选择器 一.五大基本选择符 1. *(通配符)*通配符选择器,经常用于css reset(样式重置),清理标签的默认样式,但现在一般不提倡直接使用*了,主要是*会匹配所有标 ...