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  • bzoj1901&zoj2112&cogs257 Dynamic Rankings(动态排名系统)

    bzoj1901&zoj2112&cogs257 Dynamic Rankings(动态排名系统)


    cogs

    zoj

    bzoj-权限


    题解

    bzoj和zoj都是骗访问量的233,我没有权限
    带修改区间k小值,看了学习了题解一次AC真开心。。。
    不带修改的是前缀和套主席树,(O(log_2n))查询,要修改只能(O(nlog_2n))乱搞。
    把外层前缀和改成树状数组,即第(i)个主席树从存(1) ~ (i)的值域改成存(i-lowbit(i)+1) ~ (i)的值域。然后修改和查询都是(O(log_2^2n))的。。。(一个(log)树状数组另一个线段树)
    查询就从两个根相减变成多个根减多个根了,和bit差不多。


    Code

    // It is made by XZZ
    #include<cstdio>
    #include<algorithm>
    #define Fname "dynrank"
    using namespace std;
    #define rep(a,b,c) for(rg int a=b;a<=c;a++)
    #define drep(a,b,c) for(rg int a=b;a>=c;a--)
    #define erep(a,b) for(rg int a=fir[b];a;a=nxt[a])
    #define il inline
    #define rg register
    #define vd void
    #define lb(o) ((o)&(-(o)))
    #define mid ((l+r)>>1)
    typedef long long ll;
    il int gi(){
        rg int x=0;rg char ch=getchar();
        while(ch<'0'||ch>'9')ch=getchar();
        while(ch>='0'&&ch<='9')x=x*10+ch-'0',ch=getchar();
        return x;
    }
    typedef struct node* point;
    point null;
    struct node{
        int data;
        point ls,rs;
        node(){ls=rs=null,data=0;}
        node(point _ls,point _rs){ls=_ls,rs=_rs,data=0;}
    };
    const int maxn=50001,maxm=10001;
    il vd copy(point&a,point b){
        if(b==null)a=null;
        else a=new node,*a=*b;
    }
    il vd Updata(point&s,point rt,int l,int r,int&pos,int num){
        copy(s,rt);
        s->data+=num;
        if(l==r)return;
        if(mid<pos)Updata(s->rs,rt->rs,mid+1,r,pos,num);
        else Updata(s->ls,rt->ls,l,mid,pos,num);
    }
    il point build(int l,int r){
        if(l==r)return new node;
        return new node(build(l,mid),build(mid+1,r));
    }
    point a[18],b[18];
    int _a,_b;
    il int Query(int l,int r,int k){
        while(l<r){
    	int res=0;
    	rep(i,1,_a)res+=a[i]->ls->data;
    	rep(i,1,_b)res-=b[i]->ls->data;
    	if(res>=k){
    	    rep(i,1,_a)a[i]=a[i]->ls;
    	    rep(i,1,_b)b[i]=b[i]->ls;
    	    r=mid;
    	}else{
    	    rep(i,1,_a)a[i]=a[i]->rs;
    	    rep(i,1,_b)b[i]=b[i]->rs;
    	    l=mid+1,k-=res;
    	}
        }return l;
    }
    il vd free(point i){if(i!=null)free(i->ls),free(i->rs),delete i;}
    il vd work(){
        int n=gi(),m=gi(),N=n;
        point root[maxn+maxm];
        int num[maxn],data[maxn+maxm];
        int A[maxm],B[maxm],K[maxm];
        char opt[3];
        rep(i,1,n)num[i]=data[i]=gi();
        rep(i,1,m){
    	scanf("%s",opt);
    	A[i]=gi(),B[i]=gi();
    	if(opt[0]=='Q')K[i]=gi();
    	else K[i]=-1,data[++N]=B[i];
        }
        sort(data+1,data+N+1);
        int tot=unique(data+1,data+N+1)-data-1;
        root[0]=build(1,tot);
        rep(i,1,n)num[i]=lower_bound(data+1,data+tot+1,num[i])-data,root[i]=root[i-1];
        rep(i,1,n)for(rg int j=i;j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[i],1);
        rep(i,1,m){
    	if(K[i]==-1){
    	    B[i]=lower_bound(data+1,data+tot+1,B[i])-data;
    	    for(rg int j=A[i];j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[A[i]],-1);
    	    num[A[i]]=B[i];
    	    for(rg int j=A[i];j<=n;j+=lb(j))Updata(root[j],root[j],1,tot,num[A[i]],1);
    	}else{
    	    _a=_b=0;
    	    for(rg int j=B[i];j;j-=lb(j))a[++_a]=root[j];
    	    for(rg int j=A[i]-1;j;j-=lb(j))b[++_b]=root[j];
    	    printf("%d
    ",data[Query(1,tot,K[i])]);
    	}
        }
    }
    int main(){
        freopen(Fname".in","r",stdin);
        freopen(Fname".out","w",stdout);
        int T=gi();
        null=new node;
        null->ls=null->rs=null;
        while(T--)work();
        return 0;
    }
    
    博主是蒟蒻,有问题请指出,谢谢!
    本博客中博文均为原创,未经博主允许请勿随意转载,谢谢。
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  • 原文地址:https://www.cnblogs.com/xzz_233/p/7477138.html
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