FZU FZOJ 2150 Fire Game( bfs过)

 Problem 2150 Fire Game

Accept: 1138    Submit: 4082
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

4

3 3

.#.

###

.#.

3 3

.#.

#.#

.#.

3 3

...

#.#

...

3 3

###

..#

#.#

 Sample Output

Case 1: 1

Case 2: -1

Case 3: 0

Case 4: 2

题意：俩熊孩子放火烧草地，初始让两块草地燃烧，问怎样点燃草地使得最少的等待时间，让所有草地都燃烧。

利用广搜特性：路径优先   来解决问题。这里我们知道，图是比较小的 1 <= n <=10, 1 <= m <=10，所以我们这里采用暴力搜索的方法来解决问题，因为题干中给出的意识是初始点为2.所以我们这里每一次都用两个点来进行BFS.（这里也要注意，也可能是一个点，假设就一块草地的情况）

这里对点的处理方法还是比较简单的，这里着重说一下怎么判断全部燃烧了。我们可以这样：先让两个点BFS之后，return的值就是最后那个点的时间。然后判断是否让所有的点都点燃了就行了：

#include<stdio.h>
#include<queue>
#include<iostream>
#include<string.h>
using namespace std;
struct zuobiao
{
    int x,y,output;//个人习惯用output代表步数，这里相当于step
}now,nex,f[102];
char a[102][102];
int vis[102][102];
int fx[4]={0,0,1,-1};
int fy[4]={-1,1,0,0};
int n,m;
int bfs(int x,int y,int xx,int yy)
{
    memset(vis,0,sizeof(vis));
    vis[x][y]=1;
    vis[xx][yy]=1;
    now.x=x;
    now.y=y;
    now.output=0;
    queue<zuobiao >s;
    s.push(now);
    now.x=xx;
    now.y=yy;
    now.output=0;
    s.push(now);//push进去两个点。
    int sum=0x3f3f3f3f;
    while(!s.empty())
    {
        now=s.front();
        s.pop();
        sum=now.output;
        for(int i=0;i<4;i++)
        {
            nex.x=now.x+fx[i];
            nex.y=now.y+fy[i];
            if(nex.x>=0&&nex.x<n&&nex.y>=0&&nex.y<m&&vis[nex.x][nex.y]==0&&a[nex.x][nex.y]=='#')
            {
                vis[nex.x][nex.y]=1;
                nex.output=now.output+1;
                s.push(nex);
            }
        }
    }
    return sum;//return最后的那个点的output（步数）
}
int main()
{
    int t;
    int kase=0;
    scanf("%d",&t);
    while(t--)
    {
        int cont=0;
        scanf("%d%d",&n,&m);
        for(int i=0;i<n;i++)
        {
            scanf("%s",a[i]);
            for(int j=0;j<m;j++)
            {
                if(a[i][j]=='#')//处理点。
                {
                    f[cont].x=i;
                    f[cont].y=j;
                    cont++;
                }
            }
        }
        int output=0x3f3f3f3f;
        for(int i=0;i<cont;i++)//n2;
        {
            for(int j=i;j<cont;j++)//注意这里要j=i。
            {
                int tmp=bfs(f[i].x,f[i].y,f[j].x,f[j].y);//两点bfs
                int ok=1;
                for(int k=0;k<n;k++)//暴力判断是否全部点燃了
                {
                    for(int l=0;l<m;l++)
                    {
                        if(a[k][l]=='#'&&vis[k][l]==0)
                        ok=0;
                    }
                }
                if(ok==1)
                output=min(output,tmp);
            }
        }
        if(output==0x3f3f3f3f)
        printf("Case %d: -1\n",++kase);
        else
        printf("Case %d: %d\n",++kase,output);
    }
}