bzoj1901:Zju2112 Dynamic Rankings

传送门

权值线段树套区间线段树的裸题,加了离散化就好了
或者也可以整体二分
代码(树套树):

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<map>
using namespace std;
void read(int &x) {
    char ch; bool ok;
    for(ok=0,ch=getchar(); !isdigit(ch); ch=getchar()) if(ch=='-') ok=1;
    for(x=0; isdigit(ch); x=x*10+ch-'0',ch=getchar()); if(ok) x=-x;
}
#define rg register
const int maxn=1e5+10;
int n,m,id,nmp[maxn*2],x[maxn],y[maxn],z[maxn],num,a[maxn],b[maxn*2],ls[maxn*20],rs[maxn*20],sum[maxn*20],rt[maxn*4];char op[maxn][3];
struct segment_tree
{
    void update(int x){sum[x]=sum[ls[x]]+sum[rs[x]];}
    void change(int &k,int l,int r,int a,int b)
    {
        if(!k)k=++id;int mid=(l+r)>>1;
        if(l==r){sum[k]+=b;return ;}
        if(a<=mid)change(ls[k],l,mid,a,b);
        else change(rs[k],mid+1,r,a,b);
        update(k);
    }
    int get(int x,int l,int r,int a,int b)
    {
        if(!x)return 0;int mid=(l+r)>>1,ans=0;
        if(a<=l&&b>=r)return sum[x];
        if(a<=mid)ans+=get(ls[x],l,mid,a,b);
        if(b>mid)ans+=get(rs[x],mid+1,r,a,b);
        return ans;
    }
}s[maxn*4];
map<int,int>mp;
void change(int x,int l,int r,int a,int b,int c)
{
    s[x].change(rt[x],1,n,a,c);
    if(l==r)return ;int mid=(l+r)>>1;
    if(b<=mid)change(x<<1,l,mid,a,b,c);
    else change(x<<1|1,mid+1,r,a,b,c);
}
int get(int x,int l,int r,int a,int b,int c)
{
    if(l==r)return l;int mid=(l+r)>>1,now=s[x<<1].get(rt[x<<1],1,n,a,b);
    if(c<=now)return get(x<<1,l,mid,a,b,c);
    else return get(x<<1|1,mid+1,r,a,b,c-now);
}
int main()
{
    read(n),read(m);num=n;
    for(rg int i=1;i<=n;i++)read(a[i]),b[i]=a[i];
    for(rg int i=1;i<=m;i++)
    {
        scanf("%s",op[i]+1),read(x[i]),read(y[i]);
        if(op[i][1]=='Q')read(z[i]);
        else b[++num]=y[i];
    }
    sort(b+1,b+num+1);int tot=0;
    for(rg int i=1;i<=num;i++)if(b[i]!=b[i-1])mp[b[i]]=++tot,nmp[tot]=b[i];
    for(rg int i=1;i<=n;i++)change(1,0,num,i,mp[a[i]],1);
    for(rg int i=1;i<=m;i++)
    {
        if(op[i][1]=='C')
        {
            change(1,0,num,x[i],mp[a[x[i]]],-1);
            a[x[i]]=y[i];
            change(1,0,num,x[i],mp[a[x[i]]],1);
        }
        else printf("%d\n",nmp[get(1,0,num,x[i],y[i],z[i])]);
    }
}
posted @ 2019-02-23 11:23  蒟蒻--lichenxi  阅读(...)  评论(...编辑  收藏