您的位置:首页 > 运维架构

[杂题] BZOJ 4437 [Cerc2015]Looping Labyrinth

2017-04-13 17:35 656 查看








三种情况分别对应

bfs提前结束

(n,0)且(0,m)都可达

存在(kn,km)可达

这题TM卡常啊

#include<cstdio>
#include<cstdlib>
#include<algorithm>
using namespace std;
typedef pair<int,int> abcd;
typedef long long ll;

inline char nc(){
static char buf[100000],*p1=buf,*p2=buf;
return p1==p2&&(p2=(p1=buf)+fread(buf,1,100000,stdin),p1==p2)?EOF:*p1++;
}
inline void read(int &x){
char c=nc(),b=1;
for (;!(c>='0' && c<='9');c=nc()) if (c=='-') b=-1;
for (x=0;c>='0' && c<='9';x=x*10+c-'0',c=nc()); x*=b;
}
inline void read(char *s){
char c=nc(); int len=-1;
for (;!(c=='.' || c=='#');c=nc());
for (;c=='.' || c=='#';s[++len]=c,c=nc()); s[++len]=0;
}

int n,m;
char Map[105][105];

const int M=2000000;
const int P=(1<<24)-1;

int hx[M+5],hy[M+5],next[M+5];
int inum,head[P+1];
inline void add(int x,int y){
int p=++inum; hx[p]=x; hy[p]=y; next[p]=head[((x<<9)^y)&P]; head[((x<<9)^y)&P]=p;
}
inline int ask(int x,int y){
for (int p=head[((x<<9)^y)&P];p;p=next[p])
if (hx[p]==x && hy[p]==y)
return 1;
return 0;
}

const int _dx[]={0,0,1,-1};
const int _dy[]={1,-1,0,0};

int l,r;
int Qx[M],Qy[M];

const int LIM=10000;
int cnt;

inline void bfs(){
l=r=-1; Qx[++r]=0,Qy[r]=0; add(0,0);
int x,y,sx,sy;
while (l<r && cnt<LIM){
x=Qx[++l],y=Qy[l],sx,sy;
for (int k=0;k<4;k++){
sx=x+_dx[k],sy=y+_dy[k];
if (!ask(sx,sy) && Map[(sx%n+n)%n][(sy%m+m)%m]=='.'){
Qx[++r]=sx,Qy[r]=sy; add(sx,sy);
if (sx%n==0 && sy%m==0) cnt++;
if (r==M-1) return;
}
}
}
}

inline ll Abs(ll x) { return x>0?x:-x; }
int flag=0; ll dx,dy;

inline void Query(int x,int y){
if (cnt<LIM && r<M-1) { printf("%s\n",ask(x,y)?"yes":"no"); return; }
if (flag) { printf("%s\n",ask((x%n+n)%n,(y%m+m)%m)?"yes":"no"); return; }
int L=-1e9,R=1e9;
while (L+5<R){
int m1=(L+L+R)/3,m2=(L+R+R)/3;
ll f1=Abs(x+dx*m1)+Abs(y+dy*m1),f2=Abs(x+dx*m2)+Abs(y+dy*m2);
if (f1<f2) R=m2; else L=m1;
}
int k=L;
for (int i=L;i<=R;i++)
if (Abs(x+dx*k)+Abs(y+dy*k)>=Abs(x+dx*i)+Abs(y+dy*i)){
if (ask(x+dx*i,y+dy*i)) return void(printf("yes\n"));
k=i;
}
printf("no\n");
}

int main(){
freopen("t.in","r",stdin);
freopen("t.out","w",stdout);
read(n); read(m);
for (int i=0;i<n;i++)
read(Map[i]);
bfs();
if (cnt>=LIM || r==M-1){
if (ask(n,0) && ask(0,m))
flag=1;
else{
dx=1LL<<40,dy=1LL<<40;
for (int i=1;i<=r;i++)
if (Qx[i]%n==0 && Qy[i]%m==0)
if (Qx[i]+Qy[i]<dx+dy)
dx=Qx[i],dy=Qy[i];
}
}
int Q,x,y;
read(Q);
while (Q--){
read(x); read(y);
Query(x,y);
}
return 0;
}
内容来自用户分享和网络整理,不保证内容的准确性,如有侵权内容,可联系管理员处理 点击这里给我发消息
标签: