BZOJ 1189 紧急疏散(二分+最大流)

求出所有人撤离的最短时间。由于每扇门只能通过一次，所以不能简单用bfs来搞。

显然答案是有单调性的，考虑二分，问题变成了判断时间x所有人能不能撤离。

考虑最大流。对于每扇门，每个时间通过的人数最多为1，所以将每扇门按时间x来拆成x个点。连边(time/i,1,t)来限制流量。

另外对于每个人m，如果能在时间t到达门d，那么连边(m,d/t,1)。再把源点和所有人连一条容量为1的边。

则可以通过判断最大流是否满流来得出所有人能不能撤离。

由于(n,m)<=20.所以可以很轻松的跑出答案。

 

# include <cstdio>
# include <cstring>
# include <cstdlib>
# include <iostream>
# include <vector>
# include <queue>
# include <stack>
# include <map>
# include <set>
# include <cmath>
# include <algorithm>
using namespace std;
# define lowbit(x) ((x)&(-x))
# define pi acos(-1.0)
# define eps 1e-9
# define MOD 1000000000
# define INF 1000000000
# define mem(a,b) memset(a,b,sizeof(a))
# define FOR(i,a,n) for(int i=a; i<=n; ++i)
# define FO(i,a,n) for(int i=a; i<n; ++i)
# define bug puts("H");
# define lch p<<1,l,mid
# define rch p<<1|1,mid+1,r
# define mp make_pair
# define pb push_back
typedef pair<int,int> PII;
typedef vector<int> VI;
# pragma comment(linker, "/STACK:1024000000,1024000000")
typedef long long LL;
int Scan() {
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
void Out(int a) {
    if(a<0) {putchar('-'); a=-a;}
    if(a>=10) Out(a/10);
    putchar(a%10+'0');
}
const int N=25;
//Code begin...

char ss[N][N];
struct Node{int vis[N][N];}node[N*N];
int pos, num, n, m, ps[4][2]={0,1,0,-1,1,0,-1,0};
queue<int>Q;
queue<PII>que;
struct Edge{int p, next, w;}edge[3200005];
int head[500005], cnt=2, s, t, vis[500005];

void add_edge(int u, int v, int w){
    edge[cnt].p=v; edge[cnt].w=w; edge[cnt].next=head[u]; head[u]=cnt++;
    edge[cnt].p=u; edge[cnt].w=0; edge[cnt].next=head[v]; head[v]=cnt++;
}
int bfs(){
    int i, v;
    mem(vis,-1);
    vis[s]=0; Q.push(s);
    while (!Q.empty()) {
        v=Q.front(); Q.pop();
        for (i=head[v]; i; i=edge[i].next) {
            if (edge[i].w>0 && vis[edge[i].p]==-1) {
                vis[edge[i].p]=vis[v] + 1;
                Q.push(edge[i].p);
            }
        }
    }
    return vis[t]!=-1;
}
int dfs(int x, int low){
    int i, a, temp=low;
    if (x==t) return low;
    for (i=head[x]; i; i=edge[i].next) {
        if (edge[i].w>0 && vis[edge[i].p]==vis[x]+1){
            a=dfs(edge[i].p,min(edge[i].w,temp));
            temp-=a; edge[i].w-=a; edge[i^1].w += a;
            if (temp==0) break;
        }
    }
    if (temp==low) vis[x]=-1;
    return low-temp;
}
void bulid(int x){
    mem(head,0); cnt=2;
    s=0; t=num+pos*x+1;
    FOR(i,1,num) add_edge(s,i,1);
    FOR(i,1,pos*x) add_edge(i+num,t,1);
    int tmp=0;
    FO(i,0,n) FO(j,0,m) if (ss[i][j]=='.') {
        ++tmp;
        FOR(k,1,pos) if (node[k].vis[i][j]&&node[k].vis[i][j]<=x) {
            FOR(l,node[k].vis[i][j],x) add_edge(tmp,(k-1)*x+l+num,1);
        }
    }
}
bool check(int x){
    bulid(x);
    int sum=0, tmp=0;
    while (bfs()) while (tmp=dfs(s,INF)) sum+=tmp;
    return sum==num;
}
int main ()
{
    scanf("%d%d",&n,&m);
    FO(i,0,n) scanf("%s",ss[i]);
    FO(i,0,n) FO(j,0,m) if (ss[i][j]=='.') ++num;
    FO(i,0,n) FO(j,0,m) if (ss[i][j]=='D') {
        ++pos;
        que.push(mp(i,j));
        while (!que.empty()) {
            PII tmp=que.front();
            int x=tmp.first, y=tmp.second; que.pop();
            FO(k,0,4) {
                int dx=x+ps[k][0], dy=y+ps[k][1];
                if (dx<0||dx>=n||dy<0||dy>=m||ss[dx][dy]!='.'||node[pos].vis[dx][dy]) continue;
                que.push(mp(dx,dy)); node[pos].vis[dx][dy]=node[pos].vis[x][y]+1;
            }
        }
    }
    int l=0, r=405, mid, flag=false;
    while (l<r) {
        mid=(l+r)>>1;
        if (check(mid)) r=mid, flag=true;
        else l=mid+1;
    }
    if (flag) printf("%d\n",r);
    else puts("impossible");
    return 0;
}