二分+网络流判定
首先处理出每个人和门间的距离
二分时间，连边时把每个门拆成mid个，一个人能在mid时间内到达，他也可以在这等一会儿，那么这mid个门之间连边
如果可以在x的时间内到达，那么x~mid之间他都可以出去，所以门向门加一连边
奉上未AC代码千万别复制
# include <bits/stdc++.h>
# define IL inline
# define RG register
# define Fill(a, b) memset(a, b, sizeof(a))
# define Copy(a, b) memcpy(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1010), __(1e6 + 10), INF(2147483647);
IL ll Read(){
RG char c = getchar(); RG ll x = 0, z = 1;
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, m, mp[30][30], cntr, cntd, dis[_][_], id[_][_];
int tcnt, tmp1[_ * 100], tmp2[__], w[__], fst[_ * 100], nxt[__], to[__], cnt, S, T, lev[_ * 100], cur[_ * 100], max_flow;
queue <int> Q;
IL void Add(RG int u, RG int v, RG int f){
w[cnt] = f; to[cnt] = v; nxt[cnt] = fst[u]; fst[u] = cnt++;
w[cnt] = 0; to[cnt] = u; nxt[cnt] = fst[v]; fst[v] = cnt++;
}
IL int Dfs(RG int u, RG int maxf){
if(u == T) return maxf;
RG int ret = 0;
for(RG int &e = cur[u]; e != -1; e = nxt[e]){
if(lev[to[e]] != lev[u] + 1 || !w[e]) continue;
RG int f = Dfs(to[e], min(w[e], maxf - ret));
ret += f; w[e ^ 1] += f; w[e] -= f;
if(ret == maxf) break;
}
return ret;
}
IL bool Bfs(){
Fill(lev, 0); lev[S] = 1; Q.push(S);
while(!Q.empty()){
RG int u = Q.front(); Q.pop();
for(RG int e = fst[u]; e != -1; e = nxt[e]){
if(lev[to[e]] || !w[e]) continue;
lev[to[e]] = lev[u] + 1;
Q.push(to[e]);
}
}
return lev[T];
}
IL int Maxflow(){ for(max_flow = 0; Bfs(); ) Copy(cur, fst), max_flow += Dfs(S, INF); return max_flow; }
struct Data{ int x, y, d; } u;
queue <Data> que;
int dx[] = {1, -1}, dy[] = {0, 0, 1, -1}, vis[_][_];
IL void BFS(RG int x, RG int y, RG int ID){
Fill(vis, 0); que.push((Data){x, y}); vis[x][y] = 1;
while(!que.empty()){
u = que.front(); que.pop();
for(RG int i = 0; i < 4; ++i){
RG int xx = u.x + dx[i], yy = u.y + dy[i];
if(xx < 1 || yy < 1 || xx > n || yy > m || mp[xx][yy] == 1 || vis[xx][yy]) continue;
vis[xx][yy] = 1;
if(mp[xx][yy] == 2) dis[ID][id[xx][yy]] = u.d + 1;
que.push((Data){xx, yy, u.d + 1});
}
}
}
IL int Check(RG int x){
Fill(fst, -1); cnt = 0; T = cntr + cntd * x + 1;
for(RG int i = 1; i <= cntr; ++i) Add(S, i, 1);
for(RG int i = 1; i <= cntd; ++i)
for(RG int j = 1; j <= x; ++j){
Add(cntr + x * (i - 1) + j, T, 1);
if(j < x) Add(cntr + x * (i - 1) + j, cntr + x * (i - 1) + j + 1, INF);
}
for(RG int i = 1; i <= cntr; ++i)
for(RG int j = 1; j <= cntd; ++j)
if(dis[i][j] && dis[i][j] <= x) Add(i, cntr + x * (j - 1) + dis[i][j], INF);
return Maxflow();
}
int main(RG int argc, RG char* argv[]){
n = Read(); m = Read();
for(RG int i = 1; i <= n; ++i){
RG char c[30]; scanf(" %s", c);
for(RG int j = 1; j <= m; ++j){
if(c[j - 1] == '.') id[i][j] = ++cntr;
else if(c[j - 1] == 'D') mp[i][j] = 2, id[i][j] = ++cntd;
else mp[i][j] = 1;
}
}
for(RG int i = 1; i <= n; ++i)
for(RG int j = 1; j <= m; ++j)
if(!mp[i][j]) BFS(i, j, id[i][j]);
RG int l = 0, r = 400, ans = -1;
while(l <= r){
RG int mid = (l + r) >> 1;
if(Check(mid) == cntr) r = mid - 1, ans = mid;
else l = mid + 1;
}
ans == -1 ? puts("impossible") : printf("%d
", ans);
return 0;
}