题意：给定一个图，然后有几个门，每个人要出去，但是每个门每个秒只能出去一个，然后问你最少时间才能全部出去。
析：初一看，应该是像搜索，但是怎么保证每个人出去的时候都不冲突呢，毕竟每个门每次只能出一个人，并不好处理，既然这样，我们可以把每个门和时间的做一个二元组，然后去对应每个人，这样的话，就是成了二分图的匹配，就能做了。
代码如下：
#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#include <list>
#include <assert.h>
#include <bitset>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a, b, sizeof a)
#define sz size()
#define pu push_up
#define pd push_down
#define cl clear()
#define all 1,n,1
#define FOR(i,x,n) for(int i = (x); i < (n); ++i)
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std; typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 1e20;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int maxm = 1e5 + 10;
const int mod = 30007;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, -1, 0, 1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c) {
return r >= 0 && r < n && c >= 0 && c < m;
} char s[15][15];
vector<P> door, peo;
int d[15][15][15][15]; void bfs(int r, int c){
d[r][c][r][c] = 0;
queue<P> q;
q.push(P(r, c)); while(!q.empty()){
P p = q.front(); q.pop();
for(int i = 0; i < 4; ++i){
int x = p.first + dr[i];
int y = p.second + dc[i];
if(!is_in(x, y) || d[r][c][x][y] <= d[r][c][p.fi][p.se] + 1 || s[x][y] != '.') continue;
d[r][c][x][y] = d[r][c][p.fi][p.se] + 1;
q.push(P(x, y));
}
}
} struct Edge{
int to, next;
};
Edge edge[maxn<<4];
int cnt, head[maxn]; void addEdge(int u, int v){
edge[cnt].to = v;
edge[cnt].next = head[u];
head[u]= cnt++;
} int match[maxn];
bool used[maxn]; bool dfs(int u){
used[u] = 1;
for(int i = head[u]; ~i; i = edge[i].next){
int v = edge[i].to, w = match[v];
if(w < 0 || !used[w] && dfs(w)){
match[u] = v;
match[v] = u;
return true;
}
}
return false;
} int main(){
int T; cin >> T;
while(T--){
scanf("%d %d", &n, &m);
door.cl; peo.cl;
for(int i = 0; i < n; ++i){
scanf("%s", s[i]);
}
ms(d, INF);
FOR(i, 0, n) for(int j = 0; j < m; ++j){
if(s[i][j] == '.') peo.push_back(P(i, j));
else if(s[i][j] == 'D'){
door.push_back(P(i, j));
bfs(i, j);
}
}
ms(head, -1); cnt = 0;
int sum = n * m, ss = door.sz * peo.sz;
FOR(i, 0, door.sz) for(int j = 0; j < peo.sz; ++j){
int tmp = d[door[i].fi][door[i].se][peo[j].fi][peo[j].se];
if(tmp == INF) continue;
for(int k = tmp; k <= sum; ++k){
addEdge((k-1)*door.sz + i, ss + j);
addEdge(ss + j, (k-1)*door.sz + i);
}
}
int ans = 0; ms(match, -1);
int res = -1;
for(int i = 0; i < ss; ++i) if(match[i] < 0){
ms(used, 0); if(dfs(i) && ++ans == peo.sz){ res = i / (int)door.sz + 1; break; }
}
if(res == -1) puts("impossible");
else printf("%d\n", res);
}
return 0;
}