E – Fire Game (两点bfs)

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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.

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

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

大约就是两点，bfs，然后选择用时最少的~

#include <iostream> #include <queue> #include <cstdio> #include <algorithm> #include <cstring> #define MAXN 20 using namespace std; struct node { int x; int y; int step; }p, q, a, b, que[1234]; int n, m; char ma[MAXN][MAXN];//存图 int v[MAXN][MAXN];//是否访问 int xx[4][2] = { 
  0, 1, 0, -1, -1, 0, 1, 0};//四个方向 int bfs() { queue<struct node> q_q;//结构体的队列 int time = 0;//记录所需要的时间 q_q.push(p); q_q.push(q); v[p.x][p.y] = 1;//标记已经访问 v[q.x][q.y] = 1; while(!q_q.empty())当不为空的时候 { a = q_q.front(); time = a.step; for(int i = 0;i<4;i++)//遍历四个方向 { b.x = a.x + xx[i][0]; b.y = a.y + xx[i][1]; b.step = a.step + 1; if(b.x>=0&&b.x<n&&b.y>=0&&b.y<m&&v[b.x][b.y]==0&&ma[b.x][b.y]=='#')//如果是草坪，并且没有被烧，在图中符合条件 { v[b.x][b.y] = 1; q_q.push(b);//入队 } } q_q.pop();//删除 } for(int i=0;i<n;i++)//遍历所有的点，看看草坪是否已经全部被烧完 { for(int j=0;j<m;j++) { if(ma[i][j]=='#'&&v[i][j]==0) return ;//没有被烧完，返回一个大一点的值，避免更新时间 } } return time;//否则，所有的草坪都已经被烧完了，返回时间 } int main() { int T; cin>>T;//组数 for(int t=1;t<=T;t++) { scanf("%d %d", &n, &m); int top = 0; for(int i=0;i<n;i++)//输入 { scanf("%s", ma[i]); } for(int i=0;i<n;i++)//遍历寻找草坪 { for(int j=0;j<m;j++) { if(ma[i][j]=='#') { p.x = i; p.y = j; p.step = 0; top++; que[top] = p; } } } if(top==1||top==2)//如果只有一块或两块草坪，输出0 printf("Case %d: 0\n", t); else { int min = ; for(int i=1;i<=top;i++)//否则，取任意两个点，然后bfs，看看是否可以烧完所有的草坪，并且找到最小值 { for(int j=i+1;j<=top;j++) { memset(v, 0, sizeof(v)); p = que[i]; q = que[j]; p.step = 0; q.step = 0; int time = bfs(); if(min>time) min = time; } } if(min==) printf("Case %d: -1\n", t); else printf("Case %d: %d\n", t, min); } } return 0; }