cdq分治（hdu 5618 Jam's problem again[陌上花开]、CQOI 2011 动态逆序对、hdu 4742 Pinball Game、hdu 4456 Crowd、[HEOI2016/TJOI2016]序列、[NOI2007]货币兑换 ）

hdu 5618 Jam's problem again

#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;
}

　　

动态逆序对

#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;
}

hdu 4742 Pinball Game

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;
}

hdu 4456 crowd

Description

　　详见

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;
}

 [HEOI2016/TJOI2016]序列

#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;
}

gcd论文！