题目链接
sb错误调了3hour+。。
bfs预处理出每个\(.\)到每个\(D\)的最短距离。
二分时间\(t\)，把每个\(D\)拆成\(t\)个点，这\(t\)个点两两连边，流量\(INF\)表示\(t\)个时刻都可以从这个\(D\)出。
然后枚举所有\(.\)，再枚举所有\(D\)，如果距离\(dis\)小于\(t\)，就从这个\(.\)向这个\(D\)的第\(dis\)个点连一条流量为\(1\)的边，表示从这个时刻开始这个\(.\)可以从这个\(D\)出。
然后求最大流，如果等于\(.\)的个数，说明此\(t\)可行，二分一下即可。

#include <cstdio>
#include <queue>
#include <cstdlib>
#include <cstring>
#define INF 2147483647
using namespace std;
const int MAXN = 300010;
const int N = 440;
const int MAXM = 200010;
char a[N][N];
int b[N][N], c[N][N];
struct point{int x, y, time;
}Now;
queue <int> q;
queue <point> Q;
struct Edge{int from, to, next, rest;
}e[MAXM];
int head[MAXN], num = 1, s, t, now, n, m, dis[MAXN];
inline void Add(int from, int to, int flow){e[++num] = (Edge){ from, to, head[from], flow }; head[from] = num;e[++num] = (Edge){ to, from, head[to], 0 }; head[to] = num;
}
inline int id(int i, int j){return (i - 1) * m + j;
}
int re(){memset(dis, 0, sizeof dis);q.push(s); dis[s] = 1;while(q.size()){now = q.front(); q.pop();for(int i = head[now]; i; i = e[i].next)if(e[i].rest && !dis[e[i].to])dis[e[i].to] = dis[now] + 1, q.push(e[i].to);}return dis[t];
}
int find(int u, int flow){if(u == t || !flow) return flow;/*if(u == 448){int xsxs = 1;}*/int sum = 0, T;for(int i = head[u]; i; i = e[i].next)if(e[i].rest && dis[e[i].to] == dis[u] + 1){T = find(e[i].to, min(flow - sum, e[i].rest));e[i].rest -= T; e[i ^ 1].rest += T; sum += T;}if(!sum) dis[u] = 0;return sum;
}
int dinic(){int ans = 0;while(re()) ans += find(s, INF);/*for(int i = 1; i <= num; ++i)if(e[i].to <= n * m && e[i].from == s)if(e[i].rest)printf("%d %d %d\n", (e[i].to - 1) / 12 + 1, e[i].to % 12 == 0 ? 12 : e[i].to % 12, e[i].to);system("pause");*/return ans;
}
int cnt, tot, l[] = {233, -1, 1, 0, 0}, r[] = {666, 0, 0, -1, 1}, vis[N][N], L, R;
void bfs(int x, int y){Q.push((point){x, y, 0});for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)vis[i][j] = 0;vis[x][y] = 1;while(Q.size()){Now = Q.front(); Q.pop();for(int i = 1; i <= 4; ++i){int X = Now.x + l[i], Y = Now.y + r[i];if(!X || !Y || X > n || Y > m || a[X][Y] == 'X' || vis[X][Y]) continue;Q.push((point){X, Y, Now.time + 1});vis[X][Y] = 1;if(b[X][Y]) c[id(x, y)][b[X][Y]] = Now.time + 1;}}
}
int ans, mid;
int main(){scanf("%d%d", &n, &m); s = 290000; t = 300001;for(int i = 1; i <= n; ++i)scanf("%s", a[i] + 1);for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)if(a[i][j] == 'D')b[i][j] = ++cnt;else if(a[i][j] == '.')++tot;for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)if(a[i][j] == '.')bfs(i, j);L = 1; R = 400; ans = -1;if(tot)while(L <= R){mid = (L + R) >> 1;memset(head, 0, sizeof head);num = 1;for(int i = 1; i <= n; ++i)for(int j = 1; j <= m; ++j)if(a[i][j] == '.'){Add(s, id(i, j), 1);for(int k = 1; k <= cnt; ++k){int d = id(i, j);if(c[d][k] && c[d][k] <= mid)Add(d, 400 + cnt * c[d][k] + k, 1);}}for(int i = 1; i <= cnt; ++i)for(int j = 1; j <= mid; ++j)Add(400 + cnt * j + i, t, 1), Add(400 + cnt * j + i, 400 + cnt * j + cnt + i, INF);if(dinic() == tot) ans = mid, R = mid - 1;else L = mid + 1;}if(ans == -1) printf("impossible\n");else printf("%d\n", ans);return 0;
}