本题大意：给出一个n * m的地，‘#’ 代表草， ‘.’代表陆地，每次选择这片地里的两片草，可选相等的草，选择的两片草初始状态为被燃状态，每一分钟被点燃的草会将身边的四连块点。问你需要对于给定的这片地最少需要多少分钟能燃烧完，燃烧不完输出 -1.
本题思路：很直观的一道题，枚举所有可能开始燃烧的点，更新最小值就行，无法燃烧完则是输出-1。
参考代码：
//要勤于思考，思考使人明智，不要总是因为一两个点没有想到就浪费时间
#include <cstdio>
#include <queue>
#include <cmath>
#include <cstring>
#include <vector>
using namespace std; struct node {
int x, y, step;
}now, Next; const int maxn = + , INF = 0x3f3f3f3f;
int n, m, t, ans, connected_block, maxstep, Case = ;
char maze[maxn][maxn];
bool vis_index[maxn][maxn];
vector <node> grass; bool check () {
for(int i = ; i < n; i ++)
for(int j = ; j < m; j ++)
if(maze[i][j] == '#' && !vis_index[i][j]) return false;
return true;
} bool useful(int x, int y) {
return (x >= && y >= && x < n && y < m && maze[x][y] == '#' && !vis_index[x][y]);
} void Init() {
grass.clear();
memset(vis_index, false, sizeof vis_index);
ans = INF, connected_block = ;
} int bfs(node n1, node n2) {
queue < node> Q;
memset(vis_index, false, sizeof vis_index);
Q.push(n1), Q.push(n2);
maxstep = ;
while(!Q.empty()) {
now = Q.front();
Q.pop();
if(vis_index[now.x][now.y]) continue;
maxstep = now.step;
vis_index[now.x][now.y] = true;
for(int dx = -; dx <= ; dx ++) {
for(int dy = -; dy <= ; dy ++) {
if(abs(dx - dy) == ) {
if(useful(now.x + dx, now.y + dy)) {
Next.x = now.x + dx, Next.y = now.y + dy, Next.step = now.step + ;
Q.push(Next);
}
}
}
}
}
return maxstep;
} int main () {
scanf("%d", &t);
while(t --) {
Init();
scanf("%d %d", &n, &m);
getchar();
for(int i = ; i < n; i ++) {
for(int j = ; j < m; j ++) {
maze[i][j] =getchar();
if(maze[i][j] == '#') {
node g;
g.x = i, g.y = j, g.step = ;
grass.push_back(g);
}
}
getchar();
}
//Find answer
for(int i = ; i < grass.size(); i ++)
for(int j = i; j < grass.size(); j ++) {
grass[i].step = , grass[j].step = ;
int temp = min(bfs(grass[i], grass[j]), ans);
if(check()) ans = min(temp, ans);
}
printf("Case %d: ", ++ Case);
if(ans == INF) printf("-1\n");
else printf("%d\n", ans);
}
return ;
}