题意，10*10的地图，有若干块草地“#”，草地可以点燃，并在一秒后点燃相邻的草地。有墙壁‘·‘阻挡。初始可以从任意两点点火。问烧完最短的时间。若烧不完输出-1.
题解：由于100的数据量，直接暴力。枚举两个起点，推入队列，然后bfs。烧完就返回深度，更新一个min值。
坑：（帮同学照bug）
return t.step+1;bfs后没有算上最后一步
ac代码
#define _CRT_SECURE_NO_WARNINGS
#include <iostream>
#include <cstdio>
#include <algorithm>
#include <string.h>
#include <queue>
#include <vector>
using namespace std;
char map[][];
int vis[][];
int T, n, m, ans, num;
struct Node {
int i, j, step;
Node(int i = , int j = , int step = ) :i(i), j(j), step(step) {}
};
int dir[][] = { { , },{ -, },{ , },{ ,- } };
bool check(int i, int j) {
if (i< || i >= n || j< || j >= m || vis[i][j] || map[i][j] == '.') return true;
return false;
}
int bfs(int i, int j, int x, int y) {
queue<Node>q;
q.push(Node(i, j, )); q.push(Node(x, y, ));
vis[i][j] = ; vis[x][y] = ;
int cnt;
if (i == x&&j == y) cnt = ;
else cnt = ;
while (!q.empty()) {
Node t = q.front();
q.pop();
for (int k = ; k<; k++) {
int di = t.i + dir[k][];
int dj = t.j + dir[k][];
if (check(di, dj)) continue;
q.push(Node(di, dj, t.step + ));
vis[di][dj] = ;
cnt++;
}
if (cnt >= num) {
//cout << i << ' ' << j << ' ' << x << ' ' << y <<' '<<t.step<< endl;
return t.step+; }
}
return -;
}
int main() {
scanf("%d", &T);
for (int t = ; t <= T; t++) {
ans = (int)1e9; num = ;
scanf("%d%d", &n, &m);
for (int i = ; i<n; i++) {
scanf("%s", map[i]);
for (int j = ; j<m; j++) if (map[i][j] == '#') num++;
}
if (num <= ) { printf("Case %d: %d\n", t, ); continue; }
for (int i = ; i<n; i++) {
for (int j = ; j<m; j++) {
if (map[i][j] != '#') continue;
for (int x = ; x<n; x++) {
for (int y = ; y<m; y++) {
if (map[x][y] != '#') continue;
memset(vis, , sizeof(vis));
int tmp = bfs(i, j, x, y);
if (tmp>) ans = min(ans, tmp);
}
}
}
}
if (ans == (int)1e9) ans = -;
printf("Case %d: %d\n", t, ans);
}
}