FZU 2150 Fire Game（BFS）

Problem 2150 Fire Game

Accept: 1302    Submit: 4569 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

43 3.#.###.#.3 3.#.#.#.#.3 3...#.#...3 3###..##.#

Sample Output

Case 1: 1Case 2: -1Case 3: 0Case 4: 2

```#include <iostream>
#include <string.h>
#include <math.h>
#include <stdio.h>
#include <stdlib.h>
#include <algorithm>
#include <queue>

using namespace std;
#define MAX 10000000
struct Node
{
int x;
int y;
int t;
Node(){};
Node(int x,int y,int t)
{
this->x=x;
this->y=y;
this->t=t;
}
};
char m[15][15];
int vis[15][15];
int dir[4][2]={{-1,0},{1,0},{0,1},{0,-1}};
int n,m1;
int t;
int ans;
int res;
void bfs(Node a,Node b)
{
queue<Node> q;
q.push(a);
q.push(b);
vis[a.x][a.y]=1;
vis[b.x][b.y]=1;
while(!q.empty())
{
Node term=q.front();
q.pop();
ans=max(ans,term.t);
bool tag=false;
for(int i=0;i<4;i++)
{
int xx=term.x+dir[i][0];
int yy=term.y+dir[i][1];
if(xx<1||xx>n||yy<1||yy>m1)
continue;
if(m[xx][yy]=='.'||vis[xx][yy])
continue;
vis[xx][yy]=1;tag=true;
q.push(Node(xx,yy,term.t+1));
}

}
}
int judge()
{
for(int i=1;i<=n;i++)
for(int j=1;j<=m1;j++)
if(m[i][j]=='#'&&!vis[i][j])
return 0;
return 1;
}
int main()
{
int t;
scanf("%d",&t);
int cas=0;
while(t--)
{
scanf("%d%d",&n,&m1);
for(int i=1;i<=n;i++)
scanf("%s",m[i]+1);
ans=0;res=MAX;
bool flag=false;
bool flag2=false;
for(int i=1;i<=n;i++)
{
for(int j=1;j<=m1;j++)
{
for(int k=1;k<=n;k++)
{
for(int p=1;p<=m1;p++)
{
if(m[i][j]=='#'&&m[k][p]=='#')
{
flag2=true;
memset(vis,0,sizeof(vis));
ans=0;
bfs(Node(i,j,0),Node(k,p,0));
if(judge())
res=min(res,ans);
}
}
}
}
}
if(res==MAX)
res=-1;
printf("Case %d: %d\n",++cas,res);
}
return 0;
}```

0 条评论

相关文章

2102: [Usaco2010 Dec]The Trough Game

2102: [Usaco2010 Dec]The Trough Game Time Limit: 10 Sec  Memory Limit: 64 MB Sub...

3787

2995

3536

HDU 1000 A + B Problem(指针版)

A + B Problem Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/3276...

2794

Socket与Http方式解析发送xml消息封装中间件jar包

最近项目代码中太多重复的编写Document,不同的接口需要不同的模板，于是重写提取公共部分打成jar包，方便各个系统统一使用~

1603

HDUOJ 2672---god is a girl 《斐波那契数》

god is a girl Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/3276...

2856

POJ 1012 Joseph

Joseph Time Limit: 1000MS Memory Limit: 10000K Total Submissions: 53862 ...

3326

聊聊storm的GraphiteStormReporter

storm-core-1.2.2-sources.jar!/org/apache/storm/metrics2/reporters/GraphiteStormR...

1441

2240