BZOJ3536 : [Usaco2014 Open]Cow Optics

枚举最后光线射到终点的方向，求出从起点出发以及从终点出发的光路，扫描线+树状数组统计交点个数即可。

注意当光路成环时，对应的两个方向应该只算一次。

时间复杂度$O(n\log n)$。

 

#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=100010;
const ll inf=1LL<<50;
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;
}
int n,i,j,k,q[N],g[N][4],dx[4]={0,0,-1,1},dy[4]={-1,1,0,0};char d[N],w[N*2];
int cb,cc,ce,cf,bit[N*6];ll f[N*6],ans;
struct P{
  ll x,y;
  P(){}
  P(ll _x,ll _y){x=_x,y=_y;}
  void read(){
    int a,b;
    ::read(a),::read(b);
    x=a,y=b;
  }
}B,a[N],b[N*2],c[N*2];
struct E{
  ll x,l,r;int t;
  E(){}
  E(ll _x,ll _l,ll _r,int _t){x=_x,l=_l,r=_r,t=_t;}
}e[N*6];
inline bool cmpx(int x,int y){return a[x].x==a[y].x?a[x].y<a[y].y:a[x].x<a[y].x;}
inline bool cmpy(int x,int y){return a[x].y==a[y].y?a[x].x<a[y].x:a[x].y<a[y].y;}
inline bool cmpe(const E&a,const E&b){return a.x==b.x?a.t>b.t:a.x<b.x;}
int work(P o,int k,P*q,int&m){
  q[m=1]=o;
  int i,j=-1;ll dis;
  for(i=0;i<=n;i++){
    if(k==0&&!(a[i].x==o.x&&a[i].y<=o.y))continue;
    if(k==1&&!(a[i].x==o.x&&a[i].y>=o.y))continue;
    if(k==2&&!(a[i].y==o.y&&a[i].x<=o.x))continue;
    if(k==3&&!(a[i].y==o.y&&a[i].x>=o.x))continue;
    ll t=1LL*abs(a[i].x-o.x)+1LL*abs(a[i].y-o.y);
    if(!t)continue;
    if(j<0||t<dis)j=i,dis=t;
  }
  while(~j){
    q[++m]=a[j];
    w[m]=k;
    if(!j)return 2;
    if(m>2&&q[m].x==q[2].x&&q[m].y==q[2].y&&w[m]==w[2])return 1;
    if(d[j]=='/')k=(k+2)&3;else k=3-k;
    o=a[j];
    j=g[j][k];
  }
  q[++m]=o;
  q[m].x+=inf*dx[k];
  q[m].y+=inf*dy[k];
  return 2;
}
inline void adda(const P&A,const P&B){
  if(A.y!=B.y)return;
  ll l=A.x,r=B.x;
  if(l>r)swap(l,r);
  l++,r--;
  if(l>r)return;
  e[++ce]=E(l,A.y,0,1);
  e[++ce]=E(r,A.y,0,-1);
  f[++cf]=A.y;
}
inline void addb(const P&A,const P&B){
  if(A.x!=B.x)return;
  ll l=A.y,r=B.y;
  if(l>r)swap(l,r);
  l++,r--;
  if(l>r)return;
  e[++ce]=E(A.x,l,r,0);
  f[++cf]=l,f[++cf]=r;
}
inline int lower(ll x){
  int l=1,r=cf,mid,t;
  while(l<=r)if(f[mid=(l+r)>>1]<=x)l=(t=mid)+1;else r=mid-1;
  return t;
}
inline void add(int x,int p){for(;x<=cf;x+=x&-x)bit[x]+=p;}
inline int ask(int x){int t=0;for(;x;x-=x&-x)t+=bit[x];return t;}
ll cal(){
  int i;
  sort(f+1,f+cf+1);
  for(i=1;i<=cf;i++)bit[i]=0;
  sort(e+1,e+ce+1,cmpe);
  ll ret=0;
  for(i=1;i<=ce;i++)if(e[i].t)add(lower(e[i].l),e[i].t);else ret+=ask(lower(e[i].r))-ask(lower(e[i].l)-1);
  return ret;
}
ll solve(int x){
  int i;
  ce=cf=0;
  for(i=1;i<cb;i++)adda(b[i],b[i+1]);
  for(i=3-x;i<cc;i++)addb(c[i],c[i+1]);
  ll ret=cal();
  ce=cf=0;
  for(i=1;i<cb;i++)addb(b[i],b[i+1]);
  for(i=3-x;i<cc;i++)adda(c[i],c[i+1]);
  return ret+cal();
}
int main(){
  read(n);
  B.read();
  for(i=1;i<=n;i++){
    char c[9];
    a[i].read();
    scanf("%s",c);
    d[i]=c[0];
    q[i]=i;
  }
  sort(q,q+n+1,cmpx);
  for(i=0;i<=n;i++)for(j=0;j<4;j++)g[i][j]=-1;
  for(i=0;i<=n;){
    for(j=i;j<=n&&a[q[i]].x==a[q[j]].x;j++);
    for(i++;i<j;i++){
      g[q[i]][0]=q[i-1];
      g[q[i-1]][1]=q[i];
    }
  }
  sort(q,q+n+1,cmpy);
  for(i=0;i<=n;){
    for(j=i;j<=n&&a[q[i]].y==a[q[j]].y;j++);
    for(i++;i<j;i++){
      g[q[i]][2]=q[i-1];
      g[q[i-1]][3]=q[i];
    }
  }
  work(P(0,0),1,b,cb);
  for(i=0;i<4;i++){
    int x=work(B,i,c,cc);
    ans+=solve(x)*x;
  }
  return printf("%lld",ans/2),0;
}