FZU：I. Fire Game

题目链接：http://www.bnuoj.com/v3/contest_show.php?cid=6453#problem/I

题目描述：

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.

输入格式：

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

输出格式：

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.

样例输入：

4
3 3
.#.
###
.#.
3 3
.#.
#.#
.#.
3 3
...
#.#
...
3 3
###
..#
#.#

样例输出：

Case 1: 1
Case 2: -1
Case 3: 0
Case 4: 2

分析思路：

开始想要枚举每一个点，看是否能将所有#搜完，但结果TLE。。。结果发现是有办法减少枚举量的，因为可以先用数组将所有的#信息存储下来，然后再枚举搜索所有的#就能找到最终答案了，初始化的时候注意初始化为INF，因为有些#有可能是不能被遍历到的，也可以初始化为-1.先找出选择两个#并遍历后最大的dist，再找出这其中最小的一个其实答案。

源代码如下：

//TLE...应该搜去只有#的区域，尽量进行有用的搜索
#include<cstdio>
#include<cstring>
#include<queue>
#include<algorithm>
using namespace std;
#define MAXN 50
#define INF 100000000struct Node {int x, y;
};
char Maze[MAXN][MAXN];
int dist[MAXN][MAXN];
int dir[4][2] = {-1, 0, 0, -1, 1, 0, 0, 1};
int N, M;
Node grass[MAXN*MAXN];int check(int dx, int dy) {int flag = 1;if (dx < 0 || dx >= N || dy < 0 || dy >= M) flag = 0;if (dist[dx][dy] != INF) flag = 0;return flag;
}
int BFS(Node sa, Node sd) {queue<Node> q;Node pre, now;now.x = sa.x; now.y = sa.y;q.push(now);dist[now.x][now.y] = 0;now.x = sd.x; now.y = sd.y;q.push(now);dist[now.x][now.y] = 0;while (!q.empty()) {pre = q.front(); q.pop();for (int i = 0; i < 4; i++) {now.x = pre.x + dir[i][0];now.y = pre.y + dir[i][1];if (check(now.x, now.y) && Maze[now.x][now.y] == '#') {q.push(now);dist[now.x][now.y] = dist[pre.x][pre.y] + 1;}}}
}
int main() {FILE *p = freopen("test.txt", "r", stdin);int T, i, j, d, s;while (~scanf("%d", &T)) {int c = 1;while (T--) {scanf("%d%d", &N, &M);for (i = 0; i < N; i++) {scanf("%s", Maze[i]);}int k = 0, tmin = INF;for (i = 0; i < N; i++) {for (j = 0; j < M; j++) {if (Maze[i][j] == '#') {grass[k].x = i;grass[k].y = j; k++;}}}for (i = 0; i < k; i++) {for (j = i; j < k; j++) {for (d = 0; d < N; d++) {for (s = 0; s < M; s++) {dist[d][s] = INF;}}BFS(grass[i], grass[j]);int tmax = 0;for (d = 0; d < k; d++) {tmax = max(tmax, dist[grass[d].x][grass[d].y]);//printf("dist = %d\n", dist[grass[d].x][grass[d].y]);}if (tmax == INF) continue;tmin = min(tmin, tmax);}}if (tmin == INF) printf("Case %d: -1\n", c++);else printf("Case %d: %d\n", c++, tmin);}}return 0;
}

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