zoukankan      html  css  js  c++  java
  • BZOJ3490 : Pa2011 Laser Pool

    与横线以及竖线的交点个数很容易求,那么只要求出横线竖线交点与运动轨迹的交点数即可。

    运动轨迹可以划分成若干条贯穿边界的斜线,对于第一条和最后一条,可以用bitset暴力统计。

    对于中间的部分,斜线都是完整的,可以FFT预处理。

    时间复杂度$O(nlog n+frac{nq}{32})$。

    #include<cstdio>
    #include<cmath>
    #include<algorithm>
    using namespace std;
    typedef unsigned int U;
    const int N=100010,M=10010,L=262150;
    int n,m,cb,ce,i,j,ab[N*2],arb[N*2],rab[N*2],rarb[N*2],g[65540];
    char a[N],b[N];
    struct Que{int x,y,vx,vy,t,ans;}e[M];
    inline int popcount(U x){return g[x>>16]+g[x&65535];}
    struct BIT{
      U v[N/32+5];
      void clr(){for(int i=0;i<=cb;i++)v[i]=0;}
      U get(int x){return v[x>>5]>>(x&31)&1;}
      void set(int x,U y){if((v[x>>5]>>(x&31)&1)^y)v[x>>5]^=1U<<(x&31);}
      void shl(int x,int y){
        int A=y>>5,B=y&31,C=(32-B)&31,D=x>>5,E=(D<<5)+31;
        for(int i=x;i<=E;i++)set(i,get(i+y));
        for(int i=D+1;i<=cb;i++){
          v[i]=v[i+A]>>B;
          if(C)v[i]|=v[i+A+1]<<C;
        }
      }
      void copy(int x,int y,const BIT&p){for(int i=x;i<=y;i++)v[i]=p.v[i];}
      void And(int x,int y,const BIT&p){for(int i=x;i<=y;i++)v[i]&=p.v[i];}
      int count(int x,int y){
        int A=x>>5,B=y>>5,C,ret=0;
        if(A==B){
          for(int i=x;i<=y;i++)if(v[A]>>(i&31)&1)ret++;
          return ret;
        }
        for(int i=A+1;i<B;i++)ret+=popcount(v[i]);
        C=(A<<5)+31;
        for(int i=x;i<=C;i++)if(v[A]>>(i&31)&1)ret++;
        C=B<<5;
        for(int i=C;i<=y;i++)if(v[B]>>(i&31)&1)ret++;
        return ret;
      }
    }bA,bB,brA,brB,tA,tB;
    inline void read(int&a){
      char c;bool f=0;a=0;
      while(!((((c=getchar())>='0')&&(c<='9'))||(c=='-')));
      if(c!='-')a=c-'0';else f=1;
      while(((c=getchar())>='0')&&(c<='9'))(a*=10)+=c-'0';
      if(f)a=-a;
    }
    namespace FFT{
    int k,j,pos[L];
    const double pi=acos(-1.0);
    struct comp{
      double r,i;comp(double _r=0,double _i=0){r=_r,i=_i;}
      comp operator+(const comp&x){return comp(r+x.r,i+x.i);}
      comp operator-(const comp&x){return comp(r-x.r,i-x.i);}
      comp operator*(const comp&x){return comp(r*x.r-i*x.i,r*x.i+i*x.r);}
    }a[L],ra[L],b[L],rb[L],c[L];
    void FFT(comp a[],int n,int t){
      for(int i=1;i<n;i++)if(i<pos[i])swap(a[i],a[pos[i]]);
      for(int d=0;(1<<d)<n;d++){
        int m=1<<d,m2=m<<1;
        double o=pi*2/m2*t;comp _w(cos(o),sin(o));
        for(int i=0;i<n;i+=m2){
          comp w(1,0);
          for(int j=0;j<m;j++){
            comp&A=a[i+j+m],&B=a[i+j],t=w*A;
            A=B-t;B=B+t;w=w*_w;
          }
        }
      }
      if(t==-1)for(int i=0;i<n;i++)a[i].r/=n;
    }
    void work(){
      for(k=1;k<=n||k<=m;k<<=1);k<<=1;
      j=__builtin_ctz(k)-1;
      for(i=0;i<k;i++)pos[i]=pos[i>>1]>>1|((i&1)<<j);
      for(i=1;i<=n;i++)a[i].r=ra[n-i+1].r=::a[i];
      for(i=1;i<=m;i++)b[i].r=rb[m-i+1].r=::b[i];
      FFT(a,k,1),FFT(ra,k,1),FFT(b,k,1),FFT(rb,k,1);
      for(i=0;i<k;i++)c[i]=a[i]*b[i];
      FFT(c,k,-1);
      for(i=1;i<=n+m;i++)ab[i]=(int)(c[i].r+0.5);
      for(i=0;i<k;i++)c[i]=a[i]*rb[i];
      FFT(c,k,-1);
      for(i=1;i<=n+m;i++)arb[i]=(int)(c[i].r+0.5);
      for(i=0;i<k;i++)c[i]=ra[i]*b[i];
      FFT(c,k,-1);
      for(i=1;i<=n+m;i++)rab[i]=(int)(c[i].r+0.5);
      for(i=0;i<k;i++)c[i]=ra[i]*rb[i];
      FFT(c,k,-1);
      for(i=1;i<=n+m;i++)rarb[i]=(int)(c[i].r+0.5);
    }
    }
    namespace Solve{
    bool a[N],b[N];
    int sa[N],sb[N];
    int cnt,idx[4][N],idy[4][N];
    int tot,st[N*8],en[N*8],v[N*8],pos[N*8],cur,q[N*8],sw[N*8];long long sl[N*8];
    struct E{int sx,sy,ex,ey,len,w,d,nxt;}f[N*8];
    inline bool check(int x,int y){return b[x]&&a[y];}
    inline int abs(int x){return x>0?x:-x;}
    inline int&getid(int x,int y,int d){
      if(y==1||y==n)return idx[d][x];
      return idy[d][y];
    }
    inline void makerev(int x){
      f[x].sx=f[x-1].ex;
      f[x].sy=f[x-1].ey;
      f[x].ex=f[x-1].sx;
      f[x].ey=f[x-1].sy;
      f[x].w=f[x-1].w;
      f[x].d=(f[x-1].d+2)&3;
    }
    inline int getnxt(int x,int y,int d){
      if(d==0){
        if(x<m&&y==n)return getid(x,y,(d+1)&3);
        if(y==n)return getid(x,y,(d+2)&3);
        return getid(x,y,(d+3)&3);
      }
      if(d==2){
        if(x>1&&y==1)return getid(x,y,(d+1)&3);
        if(y==1)return getid(x,y,(d+2)&3);
        return getid(x,y,(d+3)&3);
      }
      if(d==1){
        if(x<m&&y==1)return getid(x,y,(d+3)&3);
        if(y==1)return getid(x,y,(d+2)&3);
        return getid(x,y,(d+1)&3);
      }
      if(x>1&&y==n)return getid(x,y,(d+3)&3);
      if(y==n)return getid(x,y,(d+2)&3);
      return getid(x,y,(d+1)&3);
    }
    inline int cal(int x,int t,int n,int*s){
      if(x+t<=n)return s[x+t]-s[x-1];
      t-=n-x;
      int ret=s[n-1]-s[x-1];
      ret+=t/(n+n-2)*(s[n]+s[n-1]-s[1]);
      t%=n+n-2;
      if(t<n)return ret+s[n]-s[n-t-1];
      return ret+s[n]-s[1]+s[t-n+2];
    }
    inline int search(int L,int r,int x){
      int l=L,t=L-1,mid;
      while(l<=r)if(sl[mid=(l+r)>>1]-sl[L-1]<=x)l=(t=mid)+1;else r=mid-1;
      return t;
    }
    inline void ask(int x,int y,int t,int p){
      int&ans=e[p].ans;
      ans=cal(x,t,m,sb)+cal(y,t,n,sa);
      int ex=m,ey=y-x+m;
      if(ey>n)ey=n,ex=x-y+n;
      int d=ex-x;
      if(t<=d){
        tA.copy(y>>5,cb,bA);
        tB.copy(x>>5,cb,bB);
        tB.shl(0,x);
        tA.shl(0,y);
        tA.And(0,t>>5,tB);
        ans-=tA.count(0,t);
        return;
      }
      if(d){
        tA.copy(y>>5,cb,bA);
        tB.copy(x>>5,cb,bB);
        tB.shl(0,x);
        tA.shl(0,y);
        tA.And(0,(d-1)>>5,tB);
        ans-=tA.count(0,d-1);
      }
      t-=d;
      int o=getnxt(ex,ey,0);
      int l=st[v[o]],r=en[v[o]];
      o=pos[o];
      int u=search(o,r,t);
      ans-=sw[u]-sw[o-1];
      t-=sl[u]-sl[o-1];
      o=u+1;
      if(o>r){
        ans-=t/(sl[r]-sl[l-1])*(sw[r]-sw[l-1]);
        t%=sl[r]-sl[l-1];
        u=search(l,r,t);
        ans-=sw[u]-sw[l-1];
        t-=sl[u]-sl[l-1];
        o=u+1;
      }
      o=q[o];
      x=f[o].sx,y=f[o].sy;
      if(f[o].d==0){
        tA.copy(y>>5,cb,bA);
        tB.copy(x>>5,cb,bB);
      }
      if(f[o].d==1){
        y=n-y+1;
        tA.copy(y>>5,cb,brA);
        tB.copy(x>>5,cb,bB);
      }
      if(f[o].d==2){
        x=m-x+1;
        y=n-y+1;
        tA.copy(y>>5,cb,brA);
        tB.copy(x>>5,cb,brB);
      }
      if(f[o].d==3){
        x=m-x+1;
        tA.copy(y>>5,cb,bA);
        tB.copy(x>>5,cb,brB);
      }
      tB.shl(0,x);
      tA.shl(0,y);
      tA.And(0,t>>5,tB);
      ans-=tA.count(0,t);
    }
    void work(int*A,int*B){
      for(i=1;i<=n;i++)sa[i]=sa[i-1]+a[i],bA.set(i,a[i]),brA.set(n-i+1,a[i]);
      for(i=1;i<=m;i++)sb[i]=sb[i-1]+b[i],bB.set(i,b[i]),brB.set(m-i+1,b[i]);
      cnt=0;
      for(i=1;i<m;i++){
        cnt++;
        f[cnt].sx=i,f[cnt].sy=1;
        f[cnt].ex=m,f[cnt].ey=m-i+1;
        if(f[cnt].ey>n)f[cnt].ey=n,f[cnt].ex=n-1+i;
        f[cnt].w=B[m-i+2];
        f[cnt].d=0;
        makerev(++cnt);
      }
      for(i=2;i<n;i++){
        cnt++;
        f[cnt].sx=1,f[cnt].sy=i;
        f[cnt].ex=m,f[cnt].ey=m-1+i;
        if(f[cnt].ey>n)f[cnt].ey=n,f[cnt].ex=n-i+1;
        f[cnt].w=B[m+i];
        f[cnt].d=0;
        makerev(++cnt);
      }
      for(i=2;i<=m;i++){
        cnt++;
        f[cnt].sx=i,f[cnt].sy=1;
        f[cnt].ex=1,f[cnt].ey=i;
        if(f[cnt].ey>n)f[cnt].ey=n,f[cnt].ex=i+1-n;
        f[cnt].w=A[i+1];
        f[cnt].d=3;
        makerev(++cnt);
      }
      for(i=2;i<n;i++){
        cnt++;
        f[cnt].sx=m,f[cnt].sy=i;
        f[cnt].ex=1,f[cnt].ey=m+i-1;
        if(f[cnt].ey>n)f[cnt].ey=n,f[cnt].ex=m+i-n;
        f[cnt].w=A[i+m];
        f[cnt].d=3;
        makerev(++cnt);
      }
      for(i=1;i<=cnt;i++){
        f[i].len=abs(f[i].sx-f[i].ex);
        f[i].w-=check(f[i].ex,f[i].ey);
        getid(f[i].sx,f[i].sy,f[i].d)=i;
      }
      for(i=1;i<=cnt;i++)f[i].nxt=getnxt(f[i].ex,f[i].ey,f[i].d);
      cur=tot=0;
      for(i=1;i<=cnt;i++)v[i]=0;
      for(i=1;i<=cnt;i++)if(!v[i]){
        st[++tot]=cur+1;
        for(j=i;!v[j];j=f[j].nxt)v[q[++cur]=j]=tot;
        en[tot]=cur;
      }
      for(i=1;i<=cnt;i++)sl[i]=sl[i-1]+f[q[i]].len,sw[i]=sw[i-1]+f[q[i]].w,pos[q[i]]=i;
    }
    }
    int main(){
      for(i=1;i<65536;i++)g[i]=g[i>>1]+(i&1);
      read(n),read(m);
      scanf("%s%s",a+1,b+1);
      for(i=1;i<=n;i++)a[i]-='0';
      for(i=1;i<=m;i++)b[i]-='0';
      read(ce);
      for(i=1;i<=ce;i++)read(e[i].x),read(e[i].y),read(e[i].vx),read(e[i].vy),read(e[i].t);
      FFT::work();
      cb=(n>m?n:m)>>5;
      for(i=1;i<=n;i++)Solve::a[i]=a[i];
      for(i=1;i<=m;i++)Solve::b[i]=b[i];
      Solve::work(ab,arb);
      for(i=1;i<=ce;i++)if(e[i].vx==1&&e[i].vy==1)Solve::ask(e[i].x,e[i].y,e[i].t,i);
      for(i=1;i<=n;i++)Solve::a[i]=a[n-i+1];
      for(i=1;i<=m;i++)Solve::b[i]=b[i];
      Solve::work(rab,rarb);
      for(i=1;i<=ce;i++)if(e[i].vx==1&&e[i].vy==-1)Solve::ask(e[i].x,n-e[i].y+1,e[i].t,i);
      for(i=1;i<=n;i++)Solve::a[i]=a[i];
      for(i=1;i<=m;i++)Solve::b[i]=b[m-i+1];
      Solve::work(arb,ab);
      for(i=1;i<=ce;i++)if(e[i].vx==-1&&e[i].vy==1)Solve::ask(m-e[i].x+1,e[i].y,e[i].t,i);
      for(i=1;i<=n;i++)Solve::a[i]=a[n-i+1];
      for(i=1;i<=m;i++)Solve::b[i]=b[m-i+1];
      Solve::work(rarb,rab);
      for(i=1;i<=ce;i++)if(e[i].vx==-1&&e[i].vy==-1)Solve::ask(m-e[i].x+1,n-e[i].y+1,e[i].t,i);
      for(i=1;i<=ce;i++)printf("%d
    ",e[i].ans);
      return 0;
    }
    

      

  • 相关阅读:
    executable binary cannot run on android marshmallow (android 6.0)
    Android std and stl support
    Android std and stl support
    (OK) static linked & dynamically linked
    (OK)(OK) cross compile quagga-0.99.21mr2.2 for android-x86 in Fedora23
    linux和STL 常用头文件及说明
    Firefox 新增容器标签:可同时登录多个用户
    Android —— API Level
    Android added new permission model for Android 6.0 (Marshmallow).
    (OK) Android adb连接VirtualBox方式
  • 原文地址:https://www.cnblogs.com/clrs97/p/6443502.html
Copyright © 2011-2022 走看看