http://acm.hdu.edu.cn/showproblem.php?pid=5126

首先，對于一個詢問，用容斥原理可以拆成8個詢問，于是詢問變成：給定一個四元組$(i,x_i,y_i,z_i)$，問滿足$j<i$，$x_j\leq x_i$，$y_j \leq y_i$，$z_j \leq z_i$的四元組$(j,x_j,y_j,z_j)$有多少個。

四維偏序。。。。。。。。。。。

用CDQ分治套CDQ分治，把四維偏序降為三維偏序。

CDQ大法好。

//http://acm.hdu.edu.cn/showproblem.php?pid=5126 #include<cstdio> #include<cstdlib> #include<iostream> #include<fstream> #include<algorithm> #include<cstring> #include<string> #include<cmath> #include<queue> #include<stack> #include<map> #include<utility> #include<set> #include<bitset> #include<vector> #include<functional> #include<deque> #include<cctype> #include<climits> #include<complex> //#include<bits/stdc++.h>適用于CF,UOJ，但不適用于pojusing namespace std;typedef long long LL; typedef double DB; typedef pair<int,int> PII; typedef complex<DB> CP;#define mmst(a,v) memset(a,v,sizeof(a)) #define mmcy(a,b) memcpy(a,b,sizeof(a)) #define fill(a,l,r,v) fill(a+l,a+r+1,v) #define re(i,a,b) for(i=(a);i<=(b);i++) #define red(i,a,b) for(i=(a);i>=(b);i--) #define ire(i,x) for(typedef(x.begin()) i=x.begin();i!=x.end();i++) #define fi first #define se second #define m_p(a,b) make_pair(a,b) #define p_b(a) push_back(a) #define SF scanf #define PF printf #define two(k) (1<<(k))template<class T>inline T sqr(T x){return x*x;} template<class T>inline void upmin(T &t,T tmp){if(t>tmp)t=tmp;} template<class T>inline void upmax(T &t,T tmp){if(t<tmp)t=tmp;}inline int sgn(DB x){if(abs(x)<1e-9)return 0;return(x>0)?1:-1;} const DB Pi=acos(-1.0);int gint(){int res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());return (neg)?-res:res; } LL gll(){LL res=0;bool neg=0;char z;for(z=getchar();z!=EOF && z!='-' && !isdigit(z);z=getchar());if(z==EOF)return 0;if(z=='-'){neg=1;z=getchar();}for(;z!=EOF && isdigit(z);res=res*10+z-'0',z=getchar());return (neg)?-res:res; }const int maxn=50000;int q,n; struct Tstar{int p,x,y,z,s;Tstar(int p=0,int x=0,int y=0,int z=0,int s=0):p(p),x(x),y(y),z(z),s(s){}}; int res[8*maxn+100];bool cmpz(Tstar a,Tstar b){return a.z<b.z;} bool cmpp(Tstar a,Tstar b){return a.p<b.p;}int nz,tree[8*maxn+100]; #define lowbit(a) ((a)&(-a)) void update(int a,int v){for(;a<=nz;a+=lowbit(a))tree[a]+=v;} int ask(int a){int res=0;for(;a>=1;a-=lowbit(a))res+=tree[a];return res;}Tstar tmp[8*maxn+100];Tstar star2[8*maxn+100]; void CDQ2(int l,int r){if(l==r)return;int i,mid=(l+r)/2;CDQ2(l,mid);CDQ2(mid+1,r);int l1=l,l2=mid+1;while(1){while(l2<=r && star2[l2].s==0)l2++;if(l2>r)break;while(l1<=mid && star2[l1].y<=star2[l2].y){if(star2[l1].s==0)update(star2[l1].z,1);l1++;}res[star2[l2].p]+=ask(star2[l2].z);l2++;}re(i,l,l1-1)if(star2[i].s==0)update(star2[i].z,-1);l1=l,l2=mid+1;re(i,l,r){if(l1>mid){tmp[i]=star2[l2++];continue;}if(l2>r){tmp[i]=star2[l1++];continue;}if(star2[l1].y<=star2[l2].y)tmp[i]=star2[l1++];else tmp[i]=star2[l2++];}re(i,l,r)star2[i]=tmp[i];}Tstar star[8*maxn+100]; void CDQ1(int l,int r){if(l==r)return;int i,mid=(l+r)/2;CDQ1(l,mid);CDQ1(mid+1,r);int l1=l,l2=mid+1,n=0;while(1){while(l1<=mid && star[l1].s!=0)l1++;while(l2<=r && star[l2].s==0)l2++;if(l1>mid && l2>r)break;if(l1>mid){star2[++n]=star[l2++];continue;}if(l2>r){star2[++n]=star[l1++];continue;}if(star[l1].x<=star[l2].x)star2[++n]=star[l1++];else star2[++n]=star[l2++];}if(n>0)CDQ2(1,n);l1=l,l2=mid+1;re(i,l,r){if(l1>mid){tmp[i]=star[l2++];continue;}if(l2>r){tmp[i]=star[l1++];continue;}if(star[l1].x<=star[l2].x)tmp[i]=star[l1++];else tmp[i]=star[l2++];}re(i,l,r)star[i]=tmp[i];}int main(){freopen("hdu5126.in","r",stdin);freopen("hdu5126.out","w",stdout);for(int Case=gint();Case;Case--){int i,j;q=gint();n=0;re(i,1,q){int A=gint();if(A==1){int x=gint(),y=gint(),z=gint();star[++n]=Tstar(n,x,y,z,0);}else{int x1=gint(),y1=gint(),z1=gint(),x2=gint(),y2=gint(),z2=gint();star[++n]=Tstar(n,x2 ,y2 ,z2 ,1 );star[++n]=Tstar(n,x2 ,y2 ,z1-1,-1);star[++n]=Tstar(n,x2 ,y1-1,z2 ,-1);star[++n]=Tstar(n,x2 ,y1-1,z1-1,1 );star[++n]=Tstar(n,x1-1,y2 ,z2 ,-1);star[++n]=Tstar(n,x1-1,y2 ,z1-1,1 );star[++n]=Tstar(n,x1-1,y1-1,z2 ,1 );star[++n]=Tstar(n,x1-1,y1-1,z1-1,-1);}}sort(star+1,star+n+1,cmpz);nz=0;int tmpz=-100;re(i,1,n)if(star[i].z==tmpz)star[i].z=nz;else tmpz=star[i].z,star[i].z=++nz;sort(star+1,star+n+1,cmpp);re(i,1,n)res[i]=0;CDQ1(1,n);sort(star+1,star+n+1,cmpp);re(i,1,n)if(star[i].s!=0){int ans=0;re(j,0,7)ans+=star[i+j].s*res[i+j];PF("%d\n",ans);i+=7;}}return 0;} View Code

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轉載于:https://www.cnblogs.com/maijing/p/4857785.html

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