Fire Game （FZU 2150）（BFS）

题解：一开始想错了，以为只要烧完就是那个答案，但是这不是最优的结果，需要每两个点都bfs一遍，找到如果能够全部烧完，找到花费时间最小的，如果不能return -1。在bfs的时候，记录答案的方法参考了一下其他大神的思路，把能烧到地方都需要能够用个二维数组dis[ ]来标记烧到这个地方时所用的时间是多少。在经过一次两点的bfs后就需要找dis[ ]中最大的那个点，因为这一定是烧完的最后一个点。最后找能烧完的答案中最小的那个就可以了。

#include<iostream>
#include<cstring>
#include<algorithm>
#include<string>
#include<cstdio>
#include<queue>
using  namespace std;
const int INF = 0x3f3f3f3f;
char gra[20][20];
int dis[20][20];
int dx[4] = {1,-1,0,0};
int dy[4] = {0,0,1,-1};
struct node
{
    int x,y;
};
int T,n,m;
int bfs(int x1, int y1, int x2, int y2)
{
    struct node now, next;
    queue<node>q;
    int tx, ty;
    memset(dis,INF,sizeof(dis));
    now.x = x1;  now.y = y1;
    q.push(now);
    now.x = x2;  now.y = y2;
    q.push(now);
    dis[x1][y1] = dis[x2][y2] = 0;
    while(!q.empty())
    {
        now = q.front();
        q.pop();
        for(int i =0; i < 4; i ++)
        {
            tx = now.x + dx[i];
            ty = now.y + dy[i];
 if(tx >= 0 && tx < n && ty >= 0 && ty < m && gra[tx][ty] == '#' && dis[tx][ty] > dis[now.x][now.y] + 1)
            {
                dis[tx][ty] = dis[now.x][now.y] + 1;
                next.x = tx;
                next.y = ty;
                q.push(next);
            }
        }
    }
    int maxn = 0;
    for(int i = 0; i < n; i ++)
    {
        for(int j = 0; j < m; j ++)
        {
            if(gra[i][j] == '#')
            {
                maxn = max(maxn,dis[i][j]);
            }
        }
    }
    return maxn;
}
int main()
{
    cin >> T;
    for (int cas = 1; cas <= T; cas++)
    {

        cin >> n >> m;
        for (int i = 0; i < n; i++)
        {
            cin >> gra[i];
        }
        int ans = INF;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                if (gra[i][j] == '#')
                    for (int k = 0; k < n; k++)
                    {
                        for (int p = 0; p < m; p++)
                        {
                            if (gra[k][p] == '#')
                            {
                                int temp = bfs(i, j, k, p);
                                ans = min(ans, temp);
                            }
                        }
                    }
            }
        }
        if (ans == INF)
            ans = -1;
        cout << "Case " << cas << ": " << ans << endl;
    }
    return 0;
}

Problem

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