【ContestHunter】【弱省胡策】【Round8】

平衡树维护凸壳/三角函数+递推+线段树

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　　官方题解：http://pan.baidu.com/s/1sjQbY8H

洛阳城里春光好

　　题目大意：（其实出题人已经写的很简短了……直接copy的-_-。sorry！）

一个平面上的n个点构成一个点集。老师会进行Q次操作，每次操作有以下两种可能：
1. 插入操作：给定两个实数x，y，向点集中加入一个坐标为(x,y)的点。
2. 查询操作：给定实数k，取点集中任一点(x,y)，求满足方程y=kx+b的b的最大值。
现在小Z想知道，对于数学老师的每次查询，符合题意的b值是多少。

　　容易发现我们要找的点是在 凸包上 上凸壳上！（一开始总在想凸包……我真是naive

　　然后凸包上的线段的斜率满足单调性，所以我们在凸包上二分（lower_bound）即可找到所求的点……

　　这题我是用两个map来实现的，一个map存凸包，另一个map存斜率。

　　这里我处理斜率的时候少考虑了一个地方：我在插入凸包中插入一个新点时（p[x]=y），是删除它左右两个点对应直线的斜率，再加入两个新的斜率，然后对凸包进行调整。然而如果p[x]这个位置原来就有一个点的话！就应该是删掉原来的点与两边的点之间线段的斜率，再加入新的斜率，然后对凸包进行调整。

　　比较开心的是get了一份平衡树维护凸包的模板~（来自这里：http://blog.csdn.net/auto_ac/article/details/10664641）

　　其实这个是上凸壳的……如果是凸包的话我们维护两个凸壳就好了，下凸壳的维护方法：将所有点的y坐标取反，然后维护一个新的“上凸壳”（对应原图的下凸壳）这两个凸壳只有第一个点和最后一个点会是相同的。

　　另外，这题用KD-Tree应该也可做。。。判一下四个角就可以了= =（思维难度和代码难度好像要更优一点，额……不过用map的话代码难度也没有很高

　　我写KDTree的Insert的时候没有把插进来的这个点赋值aaarticlea/png;base64,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" 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 //Round8 A
 #include<vector>
 #include<map>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
     for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
     return r*v;
 }
 const int N=1e5+;
 /*******************template********************/
 //#define debug

 int n,m;
 double ans=;
 typedef map<double,double> mii;
 typedef map<double,double>::iterator iter;
 #define X first
 #define Y second
 double cross(iter o,iter a,iter b){
     return (a->X - o->X) * (b->Y - o->Y) -
            (a->Y - o->Y) * (b->X - o->X);
 }
 mii up,slop;

 inline double slope(iter a,iter b){
     return (b->Y - a->Y)/(b->X - a->X);
 }
 bool inside(mii &p,double x,double y){
     if (!p.size()) return ;
     if (x<p.begin()->X || x>p.rbegin()->X) return ;
     if (p.count(x)) return y<=p[x];
     p[x]=y;
     iter cur = p.lower_bound(x),i,j;
     i=j=cur; i--; j++;
     bool ret=cross(i,cur,j)>=;
     p.erase(cur);
     return ret;
 }
 void add(mii &p,double x,double y){
     if (inside(p,x,y)) return;
     if (p.count(x)){
         iter cur=p.lower_bound(x),i=cur,j=cur;
         i--;j++;
         if (cur!=p.begin()) slop.erase(slope(i,cur));
         if (j!=p.end()) slop.erase(slope(cur,j));
     }
     p[x]=y;
     iter cur=p.lower_bound(x),i=cur,j=cur;
     i--; j++;
     if (cur!=p.begin() && j!=p.end()) slop.erase(slope(i,j));
     if (cur!=p.begin()) slop[slope(i,cur)]=cur->X;
     if (j!=p.end()) slop[slope(cur,j)]=j->X;
     for(i=cur,i--,j=i,j--;i!=p.begin() && cur!=p.begin();i=j--)
         if (cross(j,i,cur)>=){
             slop.erase(slope(j,i));
             slop.erase(slope(i,cur));
             slop[slope(j,cur)]=cur->X;
             p.erase(i);
         }
         else break;
     for(i=cur,i++,j=i,j++;i!=p.end() && j!=p.end();i=j++)
         if (cross(cur,i,j)>=){
             slop.erase(slope(cur,i));
             slop.erase(slope(i,j));
             slop[slope(cur,j)]=j->X;
             p.erase(i);
         }
         else break;
 }
 double query(double k){
     double x,y;
     iter it=slop.lower_bound(k);
     if (it==slop.end()) x=up.begin()->X,y=up.begin()->Y;
     else x=it->Y,y=up[x];
 //    printf("query k=%f tmpk=%f x=%f y=%f\n",k,it->X,x,y);
     return y-k*x;
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("A.in","r",stdin);
     freopen("A.out","w",stdout);
 #endif
     int type=getint();
     double x,y,k;
     n=getint(); m=getint();
     F(i,,n){
         scanf("%lf%lf",&x,&y);
         add(up,x,y);
 #ifdef debug
 //    for(iter it=up.begin();it!=up.end();it++)
 //        printf("%f %f\n",it->X,it->Y);
 //    puts("");
 #endif
     }
 #ifdef debug
     puts("up tu ke:");
     for(iter it=up.begin();it!=up.end();it++)
         printf("%f %f\n",it->X,it->Y);
     puts("");
     puts("xie lv & x :");
     for(iter it=slop.begin();it!=slop.end();it++)
         printf("%f %f\n",it->X,it->Y);
     puts("");
 #endif
     char cmd[];
     double ans=;
     F(i,,m){
         scanf("%s",cmd);
         if (cmd[]=='i'){
             scanf("%lf%lf",&x,&y);
             if (type==) x=x/ans,y=y/ans;
             add(up,x,y);
         }else{
             scanf("%lf",&k);
             if (type==) k=k/ans;
             printf("%.3f\n",ans=query(k));
         }
     }
     return ;
 }

(map维护凸包)

 //Round8 A
 #include<vector>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
     for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
     return r*v;
 }
 const int N=2e5+;
 const double INF=1e30;
 /*******************template********************/

 int n,m,D,p[N<<],cnt,tot,root;
 struct node{
     double d[],mx[],mn[];
     int l,r,D,size;
     double& operator [] (int x){return d[x];}
 }t[N<<],now;
 bool cmp(int x,int y){return t[x][D]<t[y][D];}

 #define L t[o].l
 #define R t[o].r
 #define mid (l+r>>1)
 inline void Push_up(int o){
     F(i,,){
         t[o].mn[i]=min(t[o][i],min(t[L].mn[i],t[R].mn[i]));
         t[o].mx[i]=max(t[o][i],max(t[L].mx[i],t[R].mx[i]));
     }
     t[o].size=t[L].size+t[R].size+;
 }
 inline int build(int l,int r,int dir){
     D=dir;
     nth_element(p+l,p+mid,p+r+,cmp);
     int o=p[mid];
     t[o].D=dir;
     L=l<mid ? build(l,mid-,dir^) : ;
     R=mid<r ? build(mid+,r,dir^) : ;
     Push_up(o);
     return o;
 }
 inline void dfs(int o){
     if (!o) return;
     dfs(L); p[++cnt]=o; dfs(R);
 }
 inline void rebuild(int &o){
     cnt=; dfs(o);
     o=build(,cnt,t[o].D);
 }
 inline void Insert(int &o,int dir){
     if (!o){
         o=++n;
         F(i,,) t[o].mn[i]=t[o].mx[i]=t[o][i]=now[i];
         t[o].D=dir; t[o].size=;
         return;
     }
     if (now[dir]<t[o][dir]){
         Insert(L,dir^); Push_up(o);
         if (t[L].size>t[o].size*0.7) rebuild(o);
     }else{
         Insert(R,dir^); Push_up(o);
         if (t[R].size>t[o].size*0.7) rebuild(o);
     }
 }

 double ans=;
 inline double getans(int o,double k){
     if (!o) return -INF;
     return t[o][]-k*t[o][];
 }
 inline double calc(int o,double k){
     if (!o) return -INF;
     double ans=max(t[o].mn[]-k*t[o].mn[],t[o].mn[]-k*t[o].mx[]);
     ans=max(ans,max(t[o].mx[]-k*t[o].mn[],t[o].mx[]-k*t[o].mx[]));
     return ans;
 }
 inline void query(int o,double k){
     if (!o) return;
     double dl=calc(L,k),dr=calc(R,k),d0=getans(o,k);
     ans=max(ans,d0);
     if (dl<dr){
         if (dr>ans && R) query(R,k);
         if (dl>ans && L) query(L,k);
     }else{
         if (dl>ans && L) query(L,k);
         if (dr>ans && R) query(R,k);
     }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("A.in","r",stdin);
     freopen("A.out","w",stdout);
 #endif
     F(i,,) t[i].mn[i]=INF,t[i].mx[i]=-INF;
     t[].size=;

     int type=getint();
     tot=n=getint(); m=getint();
     F(i,,n) scanf("%lf%lf",&t[i][],&t[i][]),p[i]=i;
     root=build(,n,);

     char cmd[]; double k;
     F(i,,m){
         scanf("%s",cmd);
         if (cmd[]=='i'){
             scanf("%lf%lf",&now[],&now[]);
             if (type==) now[]/=ans,now[]/=ans;
             Insert(root,);
         }else{
             scanf("%lf",&k);
             if (type==) k/=ans;
             ans=-INF;
             query(root,k);
             printf("%.3f\n",ans);
         }
     }
     return ;
 }

(KDTree)

序列

　　一道很棒的题目，具体过程戳官方题解吧……

　　一个神奇的转化是对每一项都乘了一个$2sin(\frac{k}{2})$，然后利用积化和差公式，将一个$\sum$求和转化成了一个可以裂项相消，$O(1)$计算的式子……

　　然后就可以离散化用线段树维护了，感觉细节处理蛮厉害的。。。蒟蒻除了模板部分都只能抄标程

　　因为搞出来sin和cos以后就可以打tag了……且标记支持合并= =

 //Round8 C
 #include<vector>
 #include<cmath>
 #include<cstdio>
 #include<cstring>
 #include<cstdlib>
 #include<iostream>
 #include<algorithm>
 #define rep(i,n) for(int i=0;i<n;++i)
 #define F(i,j,n) for(int i=j;i<=n;++i)
 #define D(i,j,n) for(int i=j;i>=n;--i)
 using namespace std;
 typedef long long LL;
 inline int getint(){
     int r=,v=; char ch=getchar();
     for(;!isdigit(ch);ch=getchar()) if (ch=='-') r=-;
     for(; isdigit(ch);ch=getchar()) v=v*-''+ch;
     return r*v;
 }
 const int N=;
 /*******************template********************/

 int n,m,num,k,c[N<<];
 double sink;

 struct ques{
     int l,r,d;
 }q[N];

 struct node{
     double si,co;
     int tag;
 }t[N<<];
 #define L (o<<1)
 #define R (o<<1|1)
 #define mid (l+r>>1)
 #define lch L,l,mid
 #define rch R,mid+1,r

 void maintain(int o){
     t[o].si=t[L].si+t[R].si;
     t[o].co=t[L].co+t[R].co;
 }
 void change(int o,int d){
     t[o].tag+=d;
     double si1=t[o].si,si2=sin(d),co1=t[o].co,co2=cos(d);
     t[o].si=si1*co2+co1*si2;
     t[o].co=co1*co2-si1*si2;
 }
 void Push_down(int o){
     if (t[o].tag){
         change(L,t[o].tag);
         change(R,t[o].tag);
         t[o].tag=;
     }
 }
 void build(int o,int l,int r){
     if (l==r){
         t[o].si=(cos(k*c[l]-k*0.5)-cos(k*(c[l+]-)+k*0.5))/(*sink);
         t[o].co=(sin(k*(c[l+]-)+k*0.5)-sin(k*c[l]-k*0.5))/(*sink);
     }else{
         build(lch); build(rch);
         maintain(o);
     }
 }
 void update(int o,int l,int r,int ql,int qr,int v){
     if (ql<=l && qr>=r) change(o,v);
     else{
         Push_down(o);
         if (ql<=mid) update(lch,ql,qr,v);
         if (qr>mid) update(rch,ql,qr,v);
         maintain(o);
     }
 }
 double query(int o,int l,int r,int ql,int qr){
     if (ql<=l && qr>=r) return t[o].si;
     else{
         Push_down(o);
         double ans=;
         if (ql<=mid) ans+=query(lch,ql,qr);
         if (qr>mid) ans+=query(rch,ql,qr);
         return ans;
     }
 }

 int main(){
 #ifndef ONLINE_JUDGE
     freopen("C.in","r",stdin);
     freopen("C.out","w",stdout);
 #endif
     n=getint(); m=getint(); k=getint();
     sink=sin(double(k)/2.0);
     int tot=;
     c[++tot]=; c[++tot]=n+;
     F(i,,m){
         q[i].l=getint(),q[i].r=getint(),q[i].d=getint();
         c[++tot]=q[i].l; c[++tot]=q[i].r+;
     }
     sort(c+,c+tot+);
     num=unique(c+,c+tot+)-c-;
     num--; build(,,num);
     F(i,,m){
         int d=q[i].d,l,r;
         l=lower_bound(c+,c+num+,q[i].l)-c;
         r=lower_bound(c+,c+num+,q[i].r+)-c-;
         if (d) update(,,num,l,r,d);
         else printf("%.10f\n",query(,,num,l,r));
     }
     return ;
 }