【BZOJ】1189: [HNOI2007]紧急疏散evacuate（二分+bfs+网络流）

#include <cstdio>
#include <cstring>
#include <cmath>
#include <string>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
using namespace std;
typedef long long ll;
#define rep(i, n) for(int i=0; i<(n); ++i)
#define for1(i,a,n) for(int i=(a);i<=(n);++i)
#define for2(i,a,n) for(int i=(a);i<(n);++i)
#define for3(i,a,n) for(int i=(a);i>=(n);--i)
#define for4(i,a,n) for(int i=(a);i>(n);--i)
#define CC(i,a) memset(i,a,sizeof(i))
#define read(a) a=getint()
#define print(a) printf("%d", a)
#define dbg(x) cout << (#x) << " = " << (x) << endl
#define error(x) (!(x)?puts("error"):0)
#define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<‘0‘||c>‘9‘; c=getchar()) if(c==‘-‘) k=-1; for(; c>=‘0‘&&c<=‘9‘; c=getchar()) r=r*10+c-‘0‘; return k*r; }

const int N=415, oo=~0u>>1;
const int dx[4]={1, -1, 0, 0}, dy[4]={0, 0, -1, 1};
int ihead[N], cnt=1;
struct dat { int next, to, from, cap; }e[N*N*2];
void add(int u, int v, int c) {
	e[++cnt].next=ihead[u]; ihead[u]=cnt; e[cnt].from=u; e[cnt].to=v; e[cnt].cap=c;
	e[++cnt].next=ihead[v]; ihead[v]=cnt; e[cnt].from=v; e[cnt].to=u; e[cnt].cap=0;
}
int p[N], d[N], gap[N], cur[N];
int isap(int s, int t, int n) {
	for1(i, 0, n) p[i]=0, d[i]=0, gap[i]=0, cur[i]=ihead[i];
	gap[0]=n; int ret=0, f, u=s, i;
	while(d[s]<n) {
		for(i=cur[u]; i; i=e[i].next) if(e[i].cap && d[e[i].to]+1==d[u]) break;
		if(i) {
			p[e[i].to]=cur[u]=i; u=e[i].to;
			if(u==t) {
				for(f=oo; u!=s; u=e[p[u]].from) f=min(f, e[p[u]].cap);
				for(u=t; u!=s; u=e[p[u]].from) e[p[u]].cap-=f, e[p[u]^1].cap+=f;
				ret+=f;
			}
		}
		else {
			if(!(--gap[d[u]])) break;
			d[u]=n; cur[u]=ihead[u];
			for(i=ihead[u]; i; i=e[i].next) if(e[i].cap && d[e[i].to]+1<d[u]) d[u]=d[e[i].to]+1;
			++gap[d[u]];
			if(u!=s) u=e[p[u]].from;
		}
	}
	return ret;
}
int dis[N][23][23], n, tot, m, a[30][30], S, T;
struct Q { int x, y; }q[N];
inline int ID(int x, int y) { return (x-1)*m+y; }
void bfs(int x, int y, int d[23][23]) {
	int front=0, tail=0;
	for1(i, 1, n) for1(j, 1, m) d[i][j]=oo;
	q[tail].x=x, q[tail].y=y; ++tail;
	d[x][y]=0;
	Q tp;
	while(front!=tail) {
		tp=q[front++];
		x=tp.x, y=tp.y;
		rep(i, 4) {
			int fx=dx[i]+x, fy=dy[i]+y;
			if(fx<1 || fy<1 || fx>n || fy>m || a[fx][fy]!=1 || d[fx][fy]!=oo) continue;
			d[fx][fy]=d[x][y]+1;
			q[tail].x=fx; q[tail].y=fy; ++tail;
		}
	}
	//for1(i, 1, n) { for1(j, 1, m) printf("%d ", d[i][j]==oo?0:d[i][j]); puts(""); } puts("");
}
void build() {
	for1(i, 1, n) for1(j, 1, m) if(a[i][j]==2)
		bfs(i, j, dis[ID(i, j)]);
}
bool check(int t) {
	for1(i, 1, T) ihead[i]=0;
	cnt=1;
	for1(i, 1, n) for1(j, 1, m) if(a[i][j]==2) {
		int id=ID(i, j);
		for1(x, 1, n) for1(y, 1, m) if(a[x][y]==1 && dis[id][x][y]<=t)
			add(ID(x, y), id, oo);
		add(id, T, t);
	}
	for1(i, 1, n) for1(j, 1, m) if(a[i][j]==1) add(S, ID(i, j), 1);
	return isap(S, T, T)==tot;
}
int main() {
	read(n); read(m); S=n*m+1, T=S+1;
	for1(i, 1, n) for1(j, 1, m) { char c=getchar(); while(c!=‘X‘&&c!=‘.‘&&c!=‘D‘) c=getchar(); if(c==‘.‘) a[i][j]=1, ++tot; if(c==‘D‘) a[i][j]=2; }
	build();
	int l=0, r=n*m+1, mid;
	while(l<=r) {
		mid=(l+r)>>1;
		if(check(mid)) r=mid-1;
		else l=mid+1;
	}
	if(!check(r+1)) puts("impossible");
	else printf("%d\n", r+1);
	return 0;
}