#include <bits/stdc++.h> #define MAXN 100010 using namespace std; int n,k,T,xx; int ans[MAXN],c[MAXN],f[MAXN]; struct Node{ int x,y,z,id; }a[100010],b[100010]; inline int read(){ char ch; bool f=false; int res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } inline bool cmp(Node x,Node y){ if (x.x==y.x){ if (x.y==y.y){ if (x.z==y.z){ return x.id<y.id; } else return x.z<y.z; } else return x.y<y.y; } else return x.x<y.x; } inline bool cmp1(Node x,Node y){ if (x.y==y.y) return x.z<y.z; else return x.y<y.y; } inline int lowbit(int x){ return x&(-x); } inline void add(int x,int y){ while (x<=MAXN){ c[x]+=y,x+=lowbit(x); } } inline int sum(int x){ int summ=0; while (x>0){ summ+=c[x],x-=lowbit(x); } return summ; } void solve(int l,int r){ if (l>=r) return; int mid=(l+r)/2; for (register int i=l;i<=r;i++){ b[i-l+1].x=0,b[i-l+1].y=a[i].y,b[i-l+1].z=a[i].z; if (i>mid) b[i-l+1].id=a[i].id; else b[i-l+1].id=-1; } sort(b+1,b+r-l+2,cmp1); for (register int i=1;i<=r-l+1;++i) if (b[i].id==-1) add(b[i].z,1); else ans[b[i].id]+=sum(b[i].z); for(register int i=1;i<=r-l+1;++i) if(b[i].id==-1) add(b[i].z,-1); solve(l,mid); solve(mid+1,r); } int main(){ //scanf("%d",&T); T=read(); while (T--){ memset(c,0,sizeof(c)); memset(ans,0,sizeof(ans)); //scanf("%d",&n); n=read(); for (register int i=1;i<=n;++i){ a[i].x=read(),a[i].y=read(),a[i].z=read(),a[i].id=i; //scanf("%d%d%d",&a[i].x,&a[i].y,&a[i].z); //a[i].id=i; } sort(a+1,a+1+n,cmp); for (register int i=n-1;i>=1;i--){ if (a[i].x==a[i+1].x&&a[i].y==a[i+1].y&&a[i].z==a[i+1].z) xx++; else xx=0; ans[a[i].id]+=xx; } solve(1,n); for (register int i=1;i<=n;++i) printf("%d ",ans[i]); /*for (register int i=1;i<=n;++i) f[ans[i]]++; for (register int i=0;i<n;++i) printf("%d ",f[i]);*/ } return 0; }
#include <bits/stdc++.h> #define MAXN 100010 using namespace std; long long n,m,sum1,x,gg; long long ans[MAXN],f[MAXN],c[MAXN]; struct Node{ long long x,y,z,id; }a[MAXN],b[MAXN]; inline long long read(){ char ch; bool f=false; long long res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } inline bool cmp(Node x,Node y){ if (x.x==y.x) return x.y<y.y; else return x.x<y.x; } inline bool cmp1(Node x,Node y){ return x.z<y.z; } inline long long lowbit(long long x){ return x&(-x); } inline void add(long long x,long long y){ while (x<=n){ c[x]+=y,x+=lowbit(x); } } inline long long sum(long long x){ long long summ=0; while (x>0){ summ+=c[x],x-=lowbit(x); } return summ; } inline void sc(long long l,long long r){ if (l>=r) return; long long mid=(l+r)/2; for (register long long i=l;i<=r;++i){ b[i-l+1]=a[i]; if (i>mid) b[i-l+1].id=1; } sort(b+1,b+r-l+2,cmp); for (register long long i=1;i<=r-l+1;++i){ if (b[i].id==0) add(b[i].y,1); else ans[b[i].z]+=sum(n)-sum(b[i].y); } for (register long long i=1;i<=r-l+1;++i) if (b[i].id==0) add(b[i].y,-1); for (register long long i=r-l+1;i>=1;--i){ if (b[i].id==0) add(b[i].y,1); else ans[b[i].z]+=sum(b[i].y); } for (register long long i=r-l+1;i>=1;--i){ if (b[i].id==0) add(b[i].y,-1); } sc(l,mid); sc(mid+1,r); } int main(){ n=read(),m=read();gg=n; for (register long long i=1;i<=n;++i){ a[i].x=i,a[i].y=read(); f[a[i].y]=i; } for (register long long i=1;i<=m;++i) a[f[read()]].z=gg--; for (register long long i=1;i<=n;i++) if (a[i].z==0) a[i].z=gg--; sort(a+1,a+n+1,cmp1); sc(1,n); for (register long long i=1;i<=n;i++) sum1+=ans[i]; for (register long long i=n;i>n-m;--i){ printf("%lld ",sum1); sum1-=ans[i]; } return 0; }
Description
给出n个三维点对(x,y,z),求三维非严格最长上升子序列长度和最长上升子序列数量
Solution
这是一个三维LIS问题,可以用CDQ解决。
先按x排序,降一维,然后剩下y、z,在y上进行CDQ分治,按y的大小用前面的更新后面的。
z方向离散化后用树状数组维护。
#include <bits/stdc++.h> #define MAXN 100010 using namespace std; const int Mo=1<<30; int ans,T,n; int dp[MAXN],fa[MAXN],q1[MAXN],q2[MAXN]; int g[MAXN<<1],p[MAXN]; long long num[MAXN],q3[MAXN],ansn; int t1; inline int read() { char ch; bool f=false; int res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } struct Node { int x,y,z; }q[MAXN]; inline bool cmp(Node x,Node y) { if (x.x!=y.x) return x.x<y.x; if (x.y!=y.y) return x.y<y.y; return x.z<y.z; } inline bool cmp1(int x,int y) { if (q[x].y!=q[y].y) return q[x].y<q[y].y; return x<y; } inline int lowbit(int x) { return x&(-x); } inline void add(int x,int v,long long y,int maxx) { while (x<=maxx) { if (v>q2[x]) q2[x]=v,q3[x]=y; else if (v==q2[x]) q3[x]+=y; x+=lowbit(x); } } inline int sum(int x,long long &nn) { int summ=0; while (x>0) { if (summ<q2[x]) summ=q2[x],nn=q3[x]; else if (q2[x]==summ) nn+=q3[x]; x-=lowbit(x); } return summ; } void sc(int t,int w) { /*if (t>w) return;*/ if (t==w) { if (dp[t]>ans) ans=dp[t],ansn=num[t]; else if (dp[t]==ans) { ansn=ansn+num[t]; } q1[t]=q[t].z; return; } int mid=(t+w)>>1,le=t,ri=mid+1,tt1=w-t+1,maxx=mid-t+1,l,h; memmove(g+t1,fa+t,tt1*sizeof(int)); for (int i=0;i<tt1;++i) if (g[t1+i]<=mid) fa[le++]=g[t1+i]; else fa[ri++]=g[t1+i]; t1+=tt1; sc(t,mid); l=t1; t1-=tt1; for (int i=1;i<=maxx;++i) q2[i]=0; for (int i=t1;i<l;++i) { if (g[i]<=mid) { h=lower_bound(q1+t,q1+t+maxx,q[g[i]].z)-(q1+t)+1; add(h,dp[g[i]],num[g[i]],maxx); continue; } h=lower_bound(q1+t,q1+t+maxx,q[g[i]].z)-(q1+t); if (h>=maxx||q1[t+h]>q[g[i]].z) h--; if (h>=0) { long long ttt=0; int tt=sum(h+1,ttt)+1; if (tt>dp[g[i]]) dp[g[i]]=tt,num[g[i]]=ttt; else if (tt==dp[g[i]]&&tt!=1) num[g[i]]+=ttt; } } sc(mid+1,w); merge(q1+t,q1+t+maxx,q1+t+maxx,q1+t+tt1,p); memmove(q1+t,p,tt1*sizeof(int)); } int main() { T=read(); while (T--) { n=read(); ansn=t1=ans=0; for (int i=1;i<=n;++i) q[i].x=read(),q[i].y=read(),q[i].z=read(); sort(q+1,q+n+1,cmp); for (int i=1;i<=n;++i) fa[i]=i; for (int i=1;i<=n;++i){ dp[i]=1; num[i]=1; } sort(fa+1,fa+n+1,cmp1); sc(1,n); printf("%d %lld ",ans,ansn%Mo); } return 0; }
Description
Solution
本来想继续写cdq的(毕竟cdq专题2333)但网上一查题解发现好像写二维树状数组的人更多。
一想反正我也不会,我这么菜,那就学这个吧。(而且看起来码量小多了)
然后...我调了一上午QWQ
这题就是将矩形旋转45度,每个询问就相当于询问一个矩阵,用容斥口胡下然后用二维树状数组维护
#ifndef _GLIBCXX_NO_ASSERT #include <cassert> #endif #include <cctype> #include <cerrno> #include <cfloat> #include <ciso646> #include <climits> #include <clocale> #include <cmath> #include <csetjmp> #include <csignal> #include <cstdarg> #include <cstddef> #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #if __cplusplus >= 201103L #include <ccomplex> #include <cfenv> #include <cinttypes> #include <cstdalign> #include <cstdbool> #include <cstdint> #include <ctgmath> #include <cwchar> #include <cwctype> #endif // C++ #include <algorithm> #include <bitset> #include <complex> #include <deque> #include <exception> #include <fstream> #include <functional> #include <iomanip> #include <ios> #include <iosfwd> #include <iostream> #include <istream> #include <iterator> #include <limits> #include <list> #include <locale> #include <map> #include <memory> #include <new> #include <numeric> #include <ostream> #include <queue> #include <set> #include <sstream> #include <stack> #include <stdexcept> #include <streambuf> #include <string> #include <typeinfo> #include <utility> #include <valarray> #include <vector> #if __cplusplus >= 201103L #include <array> #include <atomic> #include <chrono> #include <condition_variable> #include <forward_list> #include <future> #include <initializer_list> #include <mutex> #include <random> #include <ratio> #include <regex> #include <scoped_allocator> #include <system_error> #include <thread> #include <tuple> #include <typeindex> #include <type_traits> #include <unordered_map> #include <unordered_set> #endif #define MAXN 4000010 using namespace std; int t[MAXN],q[MAXN],p[MAXN],X[MAXN],Y[MAXN],Z[MAXN]; int k,n,m,ma; inline int read() { char ch; bool f=false; int res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } inline int find (int x) { return lower_bound(q+1,q+k,x)-q; } inline int lowbit(int x) { return x&(-x); } inline void Hash(int x,int y) { int yy=y; while (x<=ma) { y=yy; while (y<=ma) { q[k++]=x*ma+y; y+=lowbit(y); } x+=lowbit(x); } } inline void add(int x,int y,int z) { int yy=y; while (x<=ma) { y=yy; while (y<=ma) { //int zz=lower_bound(q+1,q+k,(x*ma+y))-q; //t[zz]+=z; t[find(x*ma+y)]+=z; //printf("%d ",y); y+=lowbit(y); } x+=lowbit(x); } } inline int sum(int x,int y) { int summ=0,yy=y; while (x) { y=yy; while (y) { int z=find(x*ma+y); if (q[z]==x*ma+y) summ+=t[z]; //printf("%d ",z); y-=lowbit(y); } x-=lowbit(x); } //printf("%d ",summ); return summ; } int main() { while (true) { n=read(); if (n==0) return 0; k=1,ma=2*n,m=read(); memset(t,0,sizeof(t)); for (int i=1;i<=m;++i){ p[i]=read(),X[i]=read(),Y[i]=read(),Z[i]=read(); if (p[i]==1) Hash(X[i]-Y[i]+n,X[i]+Y[i]); } sort(q+1,q+k); k=unique(q+1,q+k)-q; for (int i=1;i<=m;++i){ if (p[i]==1) add(X[i]-Y[i]+n,X[i]+Y[i],Z[i]); else { int a=max(1,X[i]-Y[i]+n-Z[i]); int b=max(1,X[i]+Y[i]-Z[i]); int c=min(ma,X[i]-Y[i]+n+Z[i]); int d=min(ma,X[i]+Y[i]+Z[i]); printf("%d ",sum(c,d)-sum(c,b-1)-sum(a-1,d)+sum(a-1,b-1)); } } } return 0; }
#include<bits/stdc++.h> #define N 100010 using namespace std; int a[N],maxx[N],minn[N],fa[N],c[N],f[N]; int num,ans,x,y,n,m; inline int read() { char ch; bool f=false; int res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } inline int lowbit(int x){ return x&(-x); } inline void add(int x,int y){ while (x<=num){ if (y) c[x]=max(c[x],y); else c[x]=0; x+=lowbit(x); } } inline int sum(int x){ int summ=0; while (x>0){ summ=max(summ,c[x]),x-=lowbit(x); } return summ; } inline bool cmp(int x,int y){ return a[x]<a[y]; } inline bool cmp1(int x,int y){ return minn[x]<minn[y]; } void sc(int l,int r){ if (l==r){ if (f[l]<1) f[l]=1; return; } int mid=(l+r)>>1; sc(l,mid); for (int i=l;i<=r;++i) fa[i]=i; sort(fa+l,fa+1+mid,cmp); sort(fa+mid+1,fa+1+r,cmp1); for (int i=mid+1,j=l;i<=r;++i){ while (j<=mid&&a[fa[j]]<=minn[fa[i]]) add(maxx[fa[j++]],f[fa[j]]); f[fa[i]]=max(f[fa[i]],sum(a[fa[i]])+1); if (i==r) for (int k=l;k<j;++k) add(maxx[fa[k]],0); } sc(mid+1,r); } int main(){ n=read(),m=read(); for (int i=1;i<=n;++i) a[i]=read(); for (int i=1;i<=n;++i) maxx[i]=minn[i]=a[i]; while (m--){ x=read(),y=read(); maxx[x]=max(maxx[x],y); minn[x]=min(minn[x],y); } for (int i=1;i<=n;++i) num=max(num,a[i]); sc(1,n); for (int i=1;i<=n;++i) ans=max(ans,f[i]); printf("%d ",ans); return 0; }
[NOI2007]货币兑换
题意不想描述,写的很清晰了。。。
Solution:
好像隐隐约约中听大佬说不止要用cdq,还要用斜率优化、splay维护....感觉好可怕
然后就搁置了两三天...(颓了两三天)
设f[i]表示前i天的最大收益。
第i天可以换成B券最大数目f[i]*(1/(Rate[i]*A[i]+B[i]))
第i天可以换成的A券最大数目f[i]*(Rate[i]/(Rate[i]*A[i]+B[i]))
第i天将第j天的券全卖掉A[i]*X(j)+B[i]*Y(j)
所以f[i]=max{f[i-1],A[i]*X(j)+B[i]*Y(j)}
则我们需要求 max p=A[i]*X(j)+B[i]*Y(j)
即我们要最大化直线方程Y(j)=-A[i]/B[i]*X(j)+p/B[j]的截距
#ifndef _GLIBCXX_NO_ASSERT
#include <cassert> #endif #include <cctype> #include <cerrno> #include <cfloat> #include <ciso646> #include <climits> #include <clocale> #include <cmath> #include <csetjmp> #include <csignal> #include <cstdarg> #include <cstddef> #include <cstdio> #include <cstdlib> #include <cstring> #include <ctime> #if __cplusplus >= 201103L #include <ccomplex> #include <cfenv> #include <cinttypes> #include <cstdalign> #include <cstdbool> #include <cstdint> #include <ctgmath> #include <cwchar> #include <cwctype> #endif // C++ #include <algorithm> #include <bitset> #include <complex> #include <deque> #include <exception> #include <fstream> #include <functional> #include <iomanip> #include <ios> #include <iosfwd> #include <iostream> #include <istream> #include <iterator> #include <limits> #include <list> #include <locale> #include <map> #include <memory> #include <new> #include <numeric> #include <ostream> #include <queue> #include <set> #include <sstream> #include <stack> #include <stdexcept> #include <streambuf> #include <string> #include <typeinfo> #include <utility> #include <valarray> #include <vector> #if __cplusplus >= 201103L #include <array> #include <atomic> #include <chrono> #include <condition_variable> #include <forward_list> #include <future> #include <initializer_list> #include <mutex> #include <random> #include <ratio> #include <regex> #include <scoped_allocator> #include <system_error> #include <thread> #include <tuple> #include <typeindex> #include <type_traits> #include <unordered_map> #include <unordered_set> #endif using namespace std; const int MAXN=100010; const double inf=1e20,Inf=1e-8; struct Node { double x,y,a,b,k,z; int num; }p[MAXN],tt[MAXN]; int a[MAXN],n,cnt; double f[MAXN]; inline bool cmp(Node x,Node y) { return x.k>y.k; } inline int read() { char ch; bool f=false; int res=0; while (((ch=getchar())<'0'||ch>'9')&&ch!='-'); if (ch=='-') f=true; else res=ch-'0'; while ((ch=getchar())>='0'&&ch<='9') res=(res<<3)+(res<<1)+ch-'0'; return f?~res+1:res; } inline double pd(int a,int b){ if (!b) return -inf; else if (fabs(p[a].x-p[b].x)<Inf) return inf; else return (p[b].y-p[a].y)/(p[b].x-p[a].x); } void sc(int t,int w){ int mid=(t+w)>>1; if (t==w){ if (f[t-1]>f[t]) f[t]=f[t-1]; p[t].y=f[t]/(p[t].a*p[t].z+p[t].b); p[t].x=p[t].y*p[t].z; return; } int j=1,cnt1=t,cnt2=mid+1; for (int i=t;i<=w;++i) if (p[i].num<=mid) tt[cnt1++]=p[i]; else tt[cnt2++]=p[i]; for (int i=t;i<=w;++i) p[i]=tt[i]; sc(t,mid); cnt=0; for (int i=t;i<=mid;++i){ while (cnt>1&&pd(a[cnt-1],a[cnt])<pd(a[cnt-1],i)+Inf) cnt--; a[++cnt]=i; } for(int i=mid+1;i<=w;i++) { while(j<cnt&&pd(a[j],a[j+1])+Inf>p[i].k) j++; f[p[i].num]=max(f[p[i].num],p[a[j]].x*p[i].a+p[a[j]].y*p[i].b); } sc(mid+1,w); int t1=t,ww1=mid+1; for(int i=t;i<=w;i++) { if(((p[t1].x<p[ww1].x||(fabs(p[t1].x-p[ww1].x)<Inf&&p[t1].y<p[ww1].y))||ww1>w)&&t1<=mid) tt[i]=p[t1++]; else tt[i]=p[ww1++]; } for(int i=t;i<=w;i++) p[i]=tt[i]; } int main(){ n=read(); scanf("%lf",&f[0]); for (int i=1;i<=n;++i){ scanf("%lf%lf%lf",&p[i].a,&p[i].b,&p[i].z); p[i].k=-p[i].a/p[i].b; p[i].num=i; } sort(p+1,p+n+1,cmp); sc(1,n); printf("%.3lf ",f[n]); return 0; }