【枚举】【二分答案】【分块答案】【BFS】【最大流】【Dinic】bzoj1189 [HNOI2007]紧急疏散evacuate

【法一】枚举Time(0~N*M):

S->'.'(1);

'D'->T(Time);

'.'->'D'(dis(用BFS预处理，注意一旦到达'D'，BFS就不能继续扩展了，注意dis的初值0x7f)<=Time ? 1 : 0);

判断是否满流;

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<queue>
using namespace std;
#define INF 2147483647
#define MAXN 411
#define MAXM 350001
int v[MAXM],cap[MAXM],en,first[MAXN],next[MAXM];
int d[MAXN],cur[MAXN];
queue<int>q;
int n,S,T,N,M;
void Init_Dinic(){memset(first,-1,sizeof(first)); en=0; S=0; n=N*M+2; T=N*M+1;}
void AddEdge(const int &U,const int &V,const int &W)
{v[en]=V; cap[en]=W; next[en]=first[U]; first[U]=en++;
v[en]=U; next[en]=first[V]; first[V]=en++;}
bool bfs()
{
memset(d,-1,sizeof(d)); q.push(S); d[S]=0;
while(!q.empty())
{
int U=q.front(); q.pop();
for(int i=first[U];i!=-1;i=next[i])
if(d[v[i]]==-1 && cap[i])
{
d[v[i]]=d[U]+1;
q.push(v[i]);
}
}
return d[T]!=-1;
}
int dfs(int U,int a)
{
if(U==T || !a) return a;
int Flow=0,f;
for(int &i=cur[U];i!=-1;i=next[i])
if(d[U]+1==d[v[i]] && (f=dfs(v[i],min(a,cap[i]))))
{
cap[i]-=f; cap[i^1]+=f;
Flow+=f; a-=f; if(!a) break;
}
if(!Flow) d[U]=-1;
return Flow;
}
int max_flow()
{
int Flow=0,tmp=0;
while(bfs())
{
memcpy(cur,first,(n+5)*sizeof(int));
while(tmp=dfs(S,INF)) Flow+=tmp;
}
return Flow;
}
int dis[22*22][22*22];
bool vis[22][22];
char map[22][22];
int num[22][22];
struct Node
{
int x,y,d;
Node(const int &a,const int &b,const int &c)
{x=a;y=b;d=c;}
Node(){}
};
const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int man[22*22],door[22*22],summ,sumd;
int main()
{
scanf("%d%d",&N,&M);
for(int i=1;i<=N;++i) scanf("%s",map[i]+1);
for(int i=1;i<=N;++i)
for(int j=1;j<=M;++j)
{
num[i][j]=++en;
if(map[i][j]=='.') man[++summ]=en;
else if(map[i][j]=='D') door[++sumd]=en;
}
memset(dis,0x7f,sizeof(dis));
queue<Node>q;
for(int i=1;i<=N;++i)
for(int j=1;j<=M;++j)
if(map[i][j]=='.')
{
memset(vis,0,sizeof(vis));
q.push(Node(i,j,0));
vis[i][j]=1;
while(!q.empty())
{
Node U=q.front(); q.pop();
for(int k=0;k<4;++k)
{
int tx=U.x+dx[k],ty=U.y+dy[k];
if(tx>=1 && tx<=N && ty>=1 && ty<=M && map[tx][ty]!='X' && (!vis[tx][ty]))
{
if(map[tx][ty]=='D')
{
dis[num[i][j]][num[tx][ty]]=U.d+1;
continue;//注意：到了门立刻离开，不能继续。
}
vis[tx][ty]=1;
q.push(Node(tx,ty,U.d+1));
}
}
}
}
for(int Time=0;Time<=N*M;++Time)
{
Init_Dinic();
for(int i=1;i<=summ;++i) AddEdge(S,man[i],1);
for(int i=1;i<=sumd;++i) AddEdge(door[i],T,Time);
for(int i=1;i<=summ;++i)
for(int j=1;j<=sumd;++j)
if(dis[man[i]][door[j]]<=Time)
AddEdge(man[i],door[j],1);
if(max_flow()==summ)
{
printf("%d\n",Time);
return 0;
}
}
puts("impossible");
return 0;
}

　　

【法二】可以二分答案，但是边界总是挂……分块答案。

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
#include<queue>
using namespace std;
#define INF 2147483647
#define MAXN 411
#define MAXM 350001
int v[MAXM],cap[MAXM],en,first[MAXN],next[MAXM];
int d[MAXN],cur[MAXN];
queue<int>q;
int n,S,T,N,M;
void Init_Dinic(){memset(first,-1,sizeof(first)); en=0; S=0; n=N*M+2; T=N*M+1;}
void AddEdge(const int &U,const int &V,const int &W)
{v[en]=V; cap[en]=W; next[en]=first[U]; first[U]=en++;
v[en]=U; next[en]=first[V]; first[V]=en++;}
bool bfs()
{
memset(d,-1,sizeof(d)); q.push(S); d[S]=0;
while(!q.empty())
{
int U=q.front(); q.pop();
for(int i=first[U];i!=-1;i=next[i])
if(d[v[i]]==-1 && cap[i])
{
d[v[i]]=d[U]+1;
q.push(v[i]);
}
}
return d[T]!=-1;
}
int dfs(int U,int a)
{
if(U==T || !a) return a;
int Flow=0,f;
for(int &i=cur[U];i!=-1;i=next[i])
if(d[U]+1==d[v[i]] && (f=dfs(v[i],min(a,cap[i]))))
{
cap[i]-=f; cap[i^1]+=f;
Flow+=f; a-=f; if(!a) break;
}
if(!Flow) d[U]=-1;
return Flow;
}
int max_flow()
{
int Flow=0,tmp=0;
while(bfs())
{
memcpy(cur,first,(n+5)*sizeof(int));
while(tmp=dfs(S,INF)) Flow+=tmp;
}
return Flow;
}
int dis[22*22][22*22];
bool vis[22][22];
char map[22][22];
int num[22][22];
struct Node
{
int x,y,d;
Node(const int &a,const int &b,const int &c)
{x=a;y=b;d=c;}
Node(){}
};
const int dx[]={0,1,0,-1},dy[]={1,0,-1,0};
int man[22*22],door[22*22],summ,sumd;
void Rebuild(const int &Time)
{
Init_Dinic();
for(int i=1;i<=summ;++i) AddEdge(S,man[i],1);
for(int i=1;i<=sumd;++i) AddEdge(door[i],T,Time);
for(int i=1;i<=summ;++i)
for(int j=1;j<=sumd;++j)
if(dis[man[i]][door[j]]<=Time)
AddEdge(man[i],door[j],1);
}
int main()
{
scanf("%d%d",&N,&M);
for(int i=1;i<=N;++i) scanf("%s",map[i]+1);
for(int i=1;i<=N;++i)
for(int j=1;j<=M;++j)
{
num[i][j]=++en;
if(map[i][j]=='.') man[++summ]=en;
else if(map[i][j]=='D') door[++sumd]=en;
}
memset(dis,0x7f,sizeof(dis));
queue<Node>q;
for(int i=1;i<=N;++i)
for(int j=1;j<=M;++j)
if(map[i][j]=='.')
{
memset(vis,0,sizeof(vis));
q.push(Node(i,j,0));
vis[i][j]=1;
while(!q.empty())
{
Node U=q.front(); q.pop();
for(int k=0;k<4;++k)
{
int tx=U.x+dx[k],ty=U.y+dy[k];
if(tx>=1 && tx<=N && ty>=1 && ty<=M && map[tx][ty]!='X' && (!vis[tx][ty]))
{
if(map[tx][ty]=='D')
{
dis[num[i][j]][num[tx][ty]]=U.d+1;
continue;//注意：到了门立刻离开，不能继续。
}
vis[tx][ty]=1;
q.push(Node(tx,ty,U.d+1));
}
}
}
}
int sz=sqrt(N*M),last=0;
for(int Time=0;last<=N*M;Time+=sz)
{
Rebuild(Time);
if(max_flow()>=summ)
{
for(int i=last+1;i<=Time;++i)
{
Rebuild(i);
if(max_flow()==summ)
{printf("%d\n",i); return 0;}
}
return 0;
} last=Time;
}
puts("impossible");
return 0;
}