[BZOJ1189][P3191][HNOI2007]紧急疏散evacuate[最大流+二分答案]

对每个门bfs，求出每个点到这个门需要的时间

二分答案，每次建图的时候在当前时间内能到达就连边，不能到达就不连边

我好像做麻烦了，不需要拆点，直接这样建图就行：

S到空地连一条容量1的边，每个空地到可到达的门连一条容量1的边，每个门到T连一条容量为时间的边

#include <bits/stdc++.h>
using namespace std;
const int inf = 0x3f3f3f3f;
const int MAXN = 81007;
int n, m, s, t, dep[MAXN], maxflow, dis[25][25], N;
bool vis[25][25];
#define Get(i, j) ((i-1) * m + j)
char mat[25][25];
struct Edge {
  int v, w, next;
} G[MAXN<<2];
int tot = 1, head[MAXN], cur[MAXN], peo;
inline void add(int u, int v, int w) {
  G[++tot] = (Edge) {v, w, head[u]};head[u] = tot;
  G[++tot] = (Edge) {u, 0, head[v]};head[v] = tot;
}
inline bool bfs(int s, int t) {
  memset(dep, 0x7f, sizeof dep);
  memcpy(cur+1, head+1, sizeof(head)-4);
  queue<int>q;
  while(!q.empty()) q.pop();
  dep[s] = 0;
  q.push(s);
  while(!q.empty()) {
    int u = q.front();
    q.pop();
    for(int i = head[u]; i; i = G[i].next) {
      int v = G[i].v, w = G[i].w;
      if (dep[v] > inf && w) {
        dep[v] = dep[u] + 1;
        if (v == t) return 1;
        q.push(v);
      }
    }
  }
  return dep[t] < inf;
}
 
int dfs(int u, int t, int limit) {
  if (u == t || !limit) return limit;
  int flow = 0, f;
  for(int i = cur[u]; i; i = G[i].next) {
    cur[u] = i;
    int v = G[i].v, w = G[i].w;
    if (dep[v] == dep[u] + 1 && (f = dfs(v, t, min(w, limit)))) {
      flow += f;
      limit -= f;
      G[i].w -= f;
      G[i^1].w += f;
      if (!limit) break;
    }
  }
  return flow;
}
 
void dinic(int s, int t) {
  while(bfs(s, t)) maxflow += dfs(s, t, inf);
}
int dx[] = {0, 1, 0, -1}, dy[] = {1, 0, -1, 0};
void BFS(int sx, int sy, int step) {
  queue<int>q;
  q.push(sx), q.push(sy);
  memset(dis, 0x3f, sizeof dis);
  dis[sx][sy] = 0;
  while(!q.empty()) {
    int x = q.front();q.pop();int y = q.front(); q.pop();
    if (dis[x][y] >= step) break;
    for(int i = 0; i < 4; ++i) {
      int xx = x + dx[i], yy = y + dy[i];
      if (mat[xx][yy] == 'X' || !(xx >= 1 && xx <= n && yy >= 1 && yy <= m) || dis[xx][yy] != inf) continue; 
      dis[xx][yy] = dis[x][y] + 1;
      add(Get(sx, sy), Get(xx, yy) + 800 + dis[xx][yy] * N, 1);
      if (mat[xx][yy] == '.') q.push(xx), q.push(yy);
    }
  }
}
inline bool check(int tim) {
  memset(head, 0, sizeof head), tot = 1, maxflow = 0;
  for(int i = 1; i <= n; ++i)
    for(int j = 1; j <= m; ++j) {
      if (mat[i][j] == '.') add(s, Get(i, j), 1), BFS(i, j, tim);
      else if (mat[i][j] == 'D') {
        for(int k = 1; k < tim; ++k) add(800 + k * N + Get(i, j), 800 + (k+1) * N + Get(i, j), inf), add(800 + k * N + Get(i, j), t, 1);
        add(800 + tim * N + Get(i, j), t, 1);
      }
    }
  dinic(s, t);
  return maxflow == peo;
}
int main(void) {
  scanf("%d %d", &n, &m);
  for(int i = 1; i <= n; ++i) scanf("%s", mat[i]+1);
  for(int i = 1; i <= n; ++i) for(int j = 1; j <= m; ++j) peo += mat[i][j] == '.';
  int l = 1, r = 201, ans = -1;
  s = 70000, t = 70001, N = Get(n, m);
  while(l <= r) {
    int mid = l+r>>1;
    if (check(mid)) r = mid-1, ans = mid;
    else l = mid+1;
  }
  if (ans == -1) puts("impossible");
  else cout << l;
  return 0;
}