FZU 2150 Fire Game

Fire Game

Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u

Submit Status Practice FZU 2150

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：给出一个m*n的图，‘#’表示草坪，‘ . ’表示空地，然后可以选择在任意的两个草坪格子点火，火每 1 s会向周围四个格子扩散，问选择那两个点使得燃烧所有的草坪花费时间最小

 

#include <iostream>
#include <cstdio>
#include <cstring>
#include <queue>
#include <cstdlib>
#include <cmath>
#include <cctype>
#define N 210
#define MAXN 0xfffffff

using namespace std;
int m, n, T;
char maps[N][N];
bool f[N][N];
int dir[][] = {,, ,, ,-, -,};

struct node
{
    int x, y;
    int step;
}a[];

bool Fire()
{
    for(int i = ; i < m; i++)
    for(int j = ; j < n; j++) {
        if(maps[i][j] == '#' && !f[i][j]) {
            return false;
        }
    }
    return true;
}

int BFS(node s, node e)
{
    int i;
    queue<node>q;
    q.push(s);
    q.push(e);
    f[s.x][s.y] = f[e.x][e.y] = true;

    while(q.size()) {
        s = q.front();
        q.pop();

        e = s;
        for(i = ; i < ; i++) {
            e.x = s.x + dir[i][];
            e.y = s.y + dir[i][];
            e.step = s.step + ;
            if(e.x >=  && e.x < m && e.y >=  && e.y < n && maps[e.x][e.y] == '#' && !f[e.x][e.y]) {
                f[e.x][e.y] = true;
                q.push(e);
            }
        }
    }
    return s.step;
}

int main()
{
    scanf("%d", &T);

    for(int k = ; k <= T; k++) {
        scanf("%d %d", &m, &n);
        int b = , ans = MAXN;
        for(int i = ; i < m; i++) { //处理图像， maps值为true为草坪
            scanf(" ");
            for(int j = ; j < n; j++) {
                scanf("%c", &maps[i][j]);
                if(maps[i][j] == '#') {
                    a[b].x = i;
                    a[b].y = j;
                    a[b++].step = ;
                }
            }
        }

        for(int i = ; i < b; i++) {
            for(int j = i; j < b; j++) {
                memset(f, , sizeof(f));
                int temp_ans = BFS(a[i], a[j]);
                if(ans > temp_ans && Fire()) {
                    ans = temp_ans;
                }
            }
        }
        if(ans == MAXN)
            printf("Case %d: -1\n", k);
        else
            printf("Case %d: %d\n", k, ans);
    }

    return ;
}