【HDOJ6681】Rikka with Cake（扫描线，线段树）

题意：给定一个n*m的平面，有k条垂直或平行的直线，问将平面分成了几个互不联通的部分

n，m<=1e9，k<=1e5

思路：

刻在DNA里的二维数点

 #include<bits/stdc++.h>
 using namespace std;
 typedef long long ll;
 typedef unsigned int uint;
 typedef unsigned long long ull;
 typedef pair<int,int> PII;
 typedef pair<ll,ll> Pll;
 typedef vector<int> VI;
 typedef vector<PII> VII;
 #define N  410000
 #define M  4100000
 #define fi first
 #define se second
 #define MP make_pair
 #define pi acos(-1)
 #define mem(a,b) memset(a,b,sizeof(a))
 #define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
 #define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
 #define lowbit(x) x&(-x)
 #define Rand (rand()*(1<<16)+rand())
 #define id(x) ((x)<=B?(x):m-n/(x)+1)
 #define ls p<<1
 #define rs p<<1|1

 const ll MOD=1e9+,inv2=(MOD+)/;
       double eps=1e-;
       int INF=1e9;
       int da[]={-,,,};
       int db[]={,,-,};

 char ch[N][];
 int t[N<<],x[N],y[N],c[N],p;

 struct arr1
 {
     int t,x,y;
 }a[N];

 bool cmp1(arr1 a,arr1 b)
 {
     return a.t<b.t;
 }

 struct arr2
 {
     int x1,x2,y;
 }b[N];

 bool cmp2(arr2 a,arr2 b)
 {
     return a.y<b.y;
 }

 int read()
 {
    int v=,f=;
    char c=getchar();
    while(c<||<c) {if(c=='-') f=-; c=getchar();}
    while(<=c&&c<=) v=(v<<)+v+v+c-,c=getchar();
    return v*f;
 }

 int lisan(int x)
 {
     int l=,r=p,last=;
     while(l<=r)
     {
         int mid=(l+r)>>;
         if(c[mid]>x) r=mid-;
         if(c[mid]==x){last=mid; r=mid-;}
         if(c[mid]<x) l=mid+;
     }
     return last;
 }

 void build(int l,int r,int p)
 {
     t[p]=;
     if(l==r) return;
     int mid=(l+r)>>;
     build(l,mid,ls);
     build(mid+,r,rs);
 }

 int query(int l,int r,int x,int y,int p)
 {
     if(x<=l&&r<=y) return t[p];
     int mid=(l+r)>>;
     int s=;
     if(x<=mid) s+=query(l,mid,x,y,ls);
     if(y>mid) s+=query(mid+,r,x,y,rs);
     return s;
 }

 void update(int l,int r,int x,int v,int p)
 {
     if(l==r)
     {
         t[p]+=v;
         return;
     }
     int mid=(l+r)>>;
     if(x<=mid) update(l,mid,x,v,ls);
      else update(mid+,r,x,v,rs);
     t[p]=t[ls]+t[rs];
 }

 int main()
 {
     //freopen("1.in","r",stdin);
     //freopen("1.out","w",stdout);

     int cas;
     scanf("%d",&cas);

     while(cas--)
     {
         int n=read(),m=read(),K=read();
         ll ans=;
         int m1=,m2=;
         p=;
         rep(i,,K)
         {
             x[i]=read(),y[i]=read();
             scanf("%s",ch[i]+);
             //if(ch[i][1]=='U'&&y[i]==) ans++;
             //if(ch[i][1]=='R'&&x[i]==1) ans++;
             c[++p]=x[i];
             c[++p]=y[i];
         }
         //printf("ans=%I64d\n",ans);
         c[++p]=n;
         c[++p]=m;
         c[++p]=;
         sort(c+,c+p+);
         rep(i,,K)
         {
             x[i]=lisan(x[i]);
             y[i]=lisan(y[i]);
         }
         n=lisan(n),m=lisan(m);
         rep(i,,K)
         {
             if(ch[i][]=='U')
             {
                 m1++;
                 a[m1].t=y[i];
                 a[m1].x=x[i];
                 a[m1].y=;
                 //a[m1].x=x[i];
                 //a[m1].y1=y[i];
                 //a[m1].y2=m;
             }
             if(ch[i][]=='D')
             {
                 m1++;
                 a[m1].t=;
                 a[m1].x=x[i];
                 a[m1].y=;

                 m1++;
                 a[m1].t=y[i]+;
                 a[m1].x=x[i];
                 a[m1].y=-;
                 //a[m1].x=x[i];
                 //a[m1].y1=1;
                 //a[m1].y2=y[i];
             }
             if(ch[i][]=='L')
             {
                 m2++;
                 b[m2].x1=;
                 b[m2].x2=x[i];
                 b[m2].y=y[i];
             }
             if(ch[i][]=='R')
             {
                 m2++;
                 b[m2].x1=x[i];
                 b[m2].x2=n;
                 b[m2].y=y[i];
             }
         }
         sort(a+,a+m1+,cmp1);
         sort(b+,b+m2+,cmp2);
         build(,p,);
         int j1=,j2=;
         rep(i,,p)
         {
             while(j1<=m1&&a[j1].t==i)
             {
                 update(,p,a[j1].x,a[j1].y,);
                 j1++;
             }
             while(j2<=m2&&b[j2].y==i)
             {
                 ans+=query(,p,b[j2].x1,b[j2].x2,);
                 j2++;
             }
         }
         printf("%I64d\n",ans+);

     }

     return ;
 }