• bzoj1901:Zju2112 Dynamic Rankings


    思路:树套树,我写了两种,一种是线段树套splay,线段树维护区间信息,splay维护第k大,一种是树状数组套权值线段树(并不是什么可持久化线段树,只不过是动态开点罢了,为什么网上一大堆题解都是可持久化线段树。。。。。明明可以直接修改的,不过可持久化线段树应该也是可以写的),树状数组维护前缀和,权值线段树维护第k大,利用那道不带修改的裸题的思想即可。

    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    #define maxn 10005
     
    int n,m;
    int a[maxn];
    char s[3];
     
    inline int read(){
        int x=0,f=1;char ch=getchar();
        for (;ch<'0'||ch>'9';ch=getchar()) if (ch=='-') f=-1;
        for (;ch>='0'&&ch<='9';ch=getchar()) x=x*10+ch-'0';
        return x*f;
    }
     
    struct splay_tree{
        int ch[200*maxn][2],fa[200*maxn],note[200*maxn],val[200*maxn],size[200*maxn],cover[200*maxn],tot;
        void clear(){
            memset(val,-1,sizeof(val));
        }
        void update(int x){
            if (!x) return;
            size[x]=size[ch[x][0]]+size[ch[x][1]]+cover[x];
        }
        void rotate(int x){
            int y=fa[x],z=fa[y],bo=note[x],bo1=note[y];
            ch[y][bo]=ch[x][bo^1],fa[ch[x][bo^1]]=y;
            ch[x][bo^1]=y,fa[y]=x;
            fa[x]=z;if (z) ch[z][bo1]=x;
            note[x]=bo1,note[y]=bo^1,note[ch[y][bo]]=bo;
            update(y);
        }
        void splay(int x){
            while (note[x]!=2){
                if (note[x]==note[fa[x]]) rotate(fa[x]);
                rotate(x);
            }
            update(x);
        }
        void insert(int &x,int v){
            int y;
            while (1){
                if (val[x]==v){cover[x]++,y=x;break;}
                y=ch[x][val[x]<v];
                if (!y){
                    y=++tot;
                    val[y]=v,size[y]=1,cover[y]=1;
                    if (x) note[y]=(val[x]<v);else note[y]=2;
                    fa[y]=x,ch[y][0]=ch[y][1]=0;
                    if (x) ch[x][note[y]]=y;
                    break;
                }
                x=y;
            }
            splay(y),x=y;
        }
        int find(int x,int v){
            splay(x);
            while (1){
                if (val[x]==v) return x;
                else if (val[x]<v) x=ch[x][1];
                else x=ch[x][0];
            }
        }
        int findk(int x,int v){
            splay(x);int ans=0;
            while (x){
                if (v<val[x]) x=ch[x][0];
                else if (v==val[x]) return ans+size[ch[x][0]]+cover[x];
                else ans+=size[ch[x][0]]+cover[x],x=ch[x][1];
            }
            return ans;
        }
        int suc(int x){
            splay(x);
            int tmp=ch[x][1];
            while (ch[tmp][0]) tmp=ch[tmp][0];
            return tmp;
        }
        void del(int &rt,int x){
            if (cover[x]>1){
                splay(x);rt=x;
                cover[x]--,size[x]--;
                return;
            }
            splay(x);int y=suc(x),z=ch[x][0];
            if (!y){
                if (!z){rt=0;return;}
                fa[z]=0,note[z]=2,rt=z;return;
            }
            fa[ch[x][1]]=0,note[ch[x][1]]=2,splay(y);
            ch[y][0]=z,fa[z]=y,update(y),rt=y;
        }
    }S;
    
    struct segment_tree{
        struct treenode{
            int root;
        }tree[6*maxn];
        void build(int p,int l,int r){
            for (int i=l;i<=r;i++) S.insert(tree[p].root,a[i]);
            if (l==r) return;
            int mid=(l+r)>>1;
            build(p<<1,l,mid),build(p<<1|1,mid+1,r);
        }
        void change(int p,int l,int r,int pos,int val){
            S.del(tree[p].root,S.find(tree[p].root,a[pos])),S.insert(tree[p].root,val);
            if (l==r) return;
            int mid=(l+r)>>1;
            if (pos<=mid) change(p<<1,l,mid,pos,val);
            else change(p<<1|1,mid+1,r,pos,val);
        }
        int query(int p,int l,int r,int x,int y,int v){
            if (x<=l && r<=y) return S.findk(tree[p].root,v);
            int mid=(l+r)>>1,ans=0;
            if (x<=mid) ans+=query(p<<1,l,mid,x,y,v);
            if (y>mid) ans+=query(p<<1|1,mid+1,r,x,y,v);
            return ans;
        }
    }T;
     
    int main(){
        n=read(),m=read();S.clear();
        for (int i=1;i<=n;i++) a[i]=read();
        T.build(1,1,n);
        while (m--){
            scanf("%s",s+1);int aa=read(),b=read();
            if (s[1]=='C') T.change(1,1,n,aa,b),a[aa]=b;
            else{
                int k=read();int l=-1e9-7,r=1e9+7,ans=0;
                while (l<=r){
                    int mid=(l+r)>>1;
                    if (T.query(1,1,n,aa,b,mid)<k) l=mid+1;
                    else r=mid-1,ans=mid;
                }
                printf("%d
    ",ans);
            }
        }
        return 0;
    }
    
    #include<iostream>
    #include<cstdio>
    #include<cstring>
    #include<algorithm>
    #include<cmath>
    using namespace std;
    #define maxn 100005
     
    int n,Q,tot,top1,top2,maxv;
    int a[maxn],t1[maxn],t2[maxn];
    char s[2];
     
    struct node{
        int val,pos;
        bool operator <(const node &a)const{return val<a.val;}
    }b[maxn],c[maxn];
     
    struct query{
        int type,a,b,c;
    }q[maxn];
     
    bool cmp(node a,node b){return a.pos<b.pos;}
     
    struct segment_tree{
        int root[maxn],treedeg;
        struct node{
            int sum,ls,rs;
        }tree[10*maxn];
        void change(int &p,int l,int r,int pos,int val){
            if (!p) p=++treedeg;int mid=(l+r)>>1;
            tree[p].sum+=val;
            if (l==r) return;
            if (pos<=mid) change(tree[p].ls,l,mid,pos,val);
            else change(tree[p].rs,mid+1,r,pos,val);
        }
    }T;
     
    void change(int i,int pos,int val){
        for (;i<=n;i+=i&(-i)) T.change(T.root[i],1,maxv,pos,val);
    }
     
    void query(int x,int y,int rank){
        top1=top2=0;
        for (int i=x;i;i-=i&(-i)) t1[++top1]=T.root[i];
        for (int i=y;i;i-=i&(-i)) t2[++top2]=T.root[i];
        int l=1,r=maxv;
        while (l<r){
            int sum=0,mid=(l+r)>>1;
            for (int i=1;i<=top1;i++) sum+=T.tree[T.tree[t1[i]].ls].sum;
            for (int i=1;i<=top2;i++) sum-=T.tree[T.tree[t2[i]].ls].sum;
            if (sum>=rank){
                for (int i=1;i<=top1;i++) t1[i]=T.tree[t1[i]].ls;
                for (int i=1;i<=top2;i++) t2[i]=T.tree[t2[i]].ls;
                r=mid;
            }
            else{
                for (int i=1;i<=top1;i++) t1[i]=T.tree[t1[i]].rs;
                for (int i=1;i<=top2;i++) t2[i]=T.tree[t2[i]].rs;
                l=mid+1,rank-=sum;
            }
        }
        printf("%d
    ",b[l].val);
    }
     
    int main(){
        scanf("%d%d",&n,&Q);
        for (int i=1;i<=n;i++) scanf("%d",&a[i]),b[i].val=a[i],b[i].pos=i;int tmp=n;
        for (int i=1;i<=Q;i++){
            scanf("%s%d%d",s+1,&q[i].a,&q[i].b),q[i].type=(s[1]=='C'?0:1);
            if (q[i].type) scanf("%d",&q[i].c);else b[++tmp].val=q[i].b,b[tmp].pos=n+i;
        }
        sort(b+1,b+tmp+1),sort(a+1,a+n+1);
        for (int i=1;i<=tmp;i++)
            if (b[i].pos>n) q[b[i].pos-n].b=i;
            else c[++tot].val=i,c[tot].pos=b[i].pos;
        sort(c+1,c+tot+1,cmp);maxv=tmp;
        for (int i=1;i<=n;i++) change(i,c[i].val,1);
        for (int i=1;i<=Q;i++){
            if (q[i].type==0){
                int pos=q[i].a,val=q[i].b;
                change(pos,c[pos].val,-1),change(pos,val,1),c[pos].val=val;
            }
            else query(q[i].b,q[i].a-1,q[i].c);
        }
        return 0;
    }
    
  • 相关阅读:
    【习题 8-9 1613】 K-Graph Oddity
    【习题 8-8 UVA
    【Hello 2018 D】Too Easy Problems
    【Hello 2018 C】Party Lemonade
    【Hello 2018 B】Christmas Spruce
    【Hello 2018 A】 Modular Exponentiation
    【习题 8-7 UVA
    【习题 8-6 UVA
    【习题 8-4 UVA
    【习题 8-3 UVA
  • 原文地址:https://www.cnblogs.com/DUXT/p/6007196.html
Copyright © 2020-2023  润新知