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  • 洛谷P4254 [JSOI2008]Blue Mary开公司(李超线段树)

    题面

    传送门

    题解

    李超线段树板子

    具体可以看这里

    //minamoto
    #include<bits/stdc++.h>
    #define R register
    #define fp(i,a,b) for(R int i=(a),I=(b)+1;i<I;++i)
    #define fd(i,a,b) for(R int i=(a),I=(b)-1;i>I;--i)
    #define go(u) for(int i=head[u],v=e[i].v;i;i=e[i].nx,v=e[i].v)
    template<class T>inline bool cmax(T&a,const T&b){return a<b?a=b,1:0;}
    using namespace std;
    char buf[1<<21],*p1=buf,*p2=buf;
    inline char getc(){return p1==p2&&(p2=(p1=buf)+fread(buf,1,1<<21,stdin),p1==p2)?EOF:*p1++;}
    int read(){
        R int res,f=1;R char ch;
        while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
        for(res=ch-'0';(ch=getc())>='0'&&ch<='9';res=res*10+ch-'0');
        return res*f;
    }
    double readdb()
    {
        R double x=0,y=0.1,f=1;R char ch;
        while((ch=getc())>'9'||ch<'0')(ch=='-')&&(f=-1);
        for(x=ch-'0';(ch=getc())>='0'&&ch<='9';x=x*10+ch-'0');
        for(ch=='.'&&(ch=getc());ch>='0'&&ch<='9';x+=(ch-'0')*y,y*=0.1,ch=getc());
        return x*f;
    }
    inline char getop(){R char ch;while((ch=getc())!='Q'&&ch!='P');return ch;}
    char sr[1<<21],z[20];int C=-1,Z=0;
    inline void Ot(){fwrite(sr,1,C+1,stdout),C=-1;}
    void print(R int x){
        if(C>1<<20)Ot();if(x<0)sr[++C]='-',x=-x;
        while(z[++Z]=x%10+48,x/=10);
        while(sr[++C]=z[Z],--Z);sr[++C]='
    ';
    }
    const int N=1e5+5;
    struct node{
    	node *lc,*rc;double b,k,lv,rv;bool flag;
    	inline void ins(R double bb,R double kk,R int l,R int r){b=bb,k=kk,lv=k*l+b,rv=k*r+b,flag=1;}
    	inline double val(const R int &x){return k*x+b;}
    }pool[N<<2],*rt;int tot;
    inline node *newnode(){return &pool[tot++];}
    int n,x,q;double b,k,bb,kk,res;char op;
    void build(node* &p,int l,int r){
    	p=newnode();if(l==r)return;
    	int mid=(l+r)>>1;
    	build(p->lc,l,mid),build(p->rc,mid+1,r);
    }
    void query(node *p,int l,int r){
    	cmax(res,p->val(x));if(l==r)return;
    	int mid=(l+r)>>1;
    	if(x<=mid)query(p->lc,l,mid);
    	else query(p->rc,mid+1,r);
    }
    void update(node *p,int l,int r){
    	if(!p->flag)return p->ins(b,k,l,r),void();
    	double lv=l*k+b,rv=r*k+b;
    	if(lv<=p->lv&&rv<=p->rv)return;
    	if(lv>p->lv&&rv>p->rv)return p->ins(b,k,l,r),void();
    	int mid=(l+r)>>1;
    	double x=(b-p->b)/(p->k-k);
    	if(lv<=p->lv){
    		if(x<=mid)bb=p->b,kk=p->k,p->ins(b,k,l,r),b=bb,k=kk,update(p->lc,l,mid);
    		else update(p->rc,mid+1,r);
    	}else{
    		if(x<=mid)update(p->lc,l,mid);
    		else bb=p->b,kk=p->k,p->ins(b,k,l,r),b=bb,k=kk,update(p->rc,mid+1,r);
    	}
    }
    int main(){
    //	freopen("testdata.in","r",stdin);
    	n=50000,q=read();
    	build(rt,1,n);
    	while(q--){
    		op=getop();
    		if(op=='Q')x=read(),res=0,query(rt,1,n),print(res/100);
    		else b=readdb(),k=readdb(),b-=k,update(rt,1,n);
    	}
    	return Ot(),0;
    }
    
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  • 原文地址:https://www.cnblogs.com/bztMinamoto/p/10620267.html
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