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
解题思路：
W代表草坪，只有草坪才能燃烧，让你给两块草坪点火，火势只能向上下左右蔓延，问把所有草坪烧完的最短时间。求最短时间又得把对整个图进行检索，典型的BFS，火向四周蔓延一次，时间加一。该题特别之处就在于是两处同时进行BFS。其实完全可以看做交替进行BFS，即你先放火，我再放火，你的火先蔓延一块草地，我再跟着蔓延一块草地，直到烧完。所以检索的时候，当第一把火燃烧到第二把火蔓延的地方会发现，悲剧了，那里的草已经被烧完了。。。。这样间接保证了时间上的同步。
AC代码：
#include <iostream>
#include <cstdio>
#include <queue>
#include <cstring>
const int maxn = 20, INF = 999999999;
using namespace std;
bool visited[maxn][maxn];
int pos, n, m, dir[4][2] = {{0,1},{1,0},{-1,0},{0,-1}}; // 定义一个二维数组代表四个方向
char map[maxn][maxn];
struct node
{
int x, y;
}p[maxn * maxn]; // 用结构体记录草坪的坐标
queue<node> point;
int min(int a,int b)
{
return a < b ? a : b;
}
bool Flag(int x,int y) // 判断该坐标是否是草坪并有无被烧过
{
if(x >= 0 && x < n && y >= 0 && y < m && map[x][y] == '#' && visited[x][y] == 0)
return true;
return false;
}
int Bfs()
{
int time = 0;
while(!point.empty())
{
int m =point.size();
while(m--)
{
node now = point.front();
point.pop();
for(int j = 0; j < 4; j++)
{
int temp_x = now.x + dir[j][0];
int temp_y = now.y + dir[j][1];
if(Flag(temp_x,temp_y))
{
node temp;
temp.x = temp_x;
temp.y = temp_y;
visited[temp.x][temp.y] = 1;
point.push(temp);
pos--;
}
}
}
time++;
}
if(pos == 0) // 判断草坪是否烧完，烧完即为0
return time - 1;
else
return INF;
}
int main()
{
int t, ans, count = 0;
scanf("%d", &t);
while(t--)
{
memset(visited, 0, sizeof(visited));
int k = 0;
ans = INF;
scanf("%d%d", &n, &m);
for(int i = 0; i < n; i++)
{
scanf("%s", map[i]);
for(int j = 0; j < m; j++)
{
if(map[i][j] == '#') // 判断该坐标是否是草坪
{
p[k].x = i;
p[k].y = j;
k++;
}
}
}
for(int i = 0; i < k - 1; i++) // 遍历所有可能的两处放火地点
for(int j = i + 1; j < k; j++)
{
memset(visited, 0, sizeof(visited));
point.push(p[i]);
point.push(p[j]);
visited[p[i].x][p[i].y] = 1;
visited[p[j].x][p[j].y] = 1; // 放火的两处标记已访问
pos = k - 2; // pos记录了还需要烧的草坪，减去了两处放火的草坪
int t = Bfs();
ans = min(ans,t); // 找出最小的燃烧时间
}
if(k == 1) // 不要漏了整块地图只有一块草坪的情况，这时候只能在同一地点放火，时间为0
printf("Case %d: 0\n", ++count);
else if(ans == INF)
printf("Case %d: -1\n", ++count);
else
printf("Case %d: %d\n", ++count, ans);
}
return 0;
}