• 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
    ",nmp[get(1,0,num,x[i],y[i],z[i])]);
    	}
    }
    
  • 相关阅读:
    synchronized关键字原理
    http几种请求格式总结
    logback配置
    docker部署nacos单机
    Diango migrate遇到问题
    pip安装ujson报错: error:Microsoft Visual C++ 14.0 is required
    vue watch监听新增属性
    git commit message规范与约束(全局安装)
    git commit message规范与约束(项目内安装)
    pip常用方法
  • 原文地址:https://www.cnblogs.com/lcxer/p/10422040.html
Copyright © 2020-2023  润新知