【bzoj3938】 Robot

http://www.lydsy.com/JudgeOnline/problem.php?id=3938 (题目链接)

题意

　　给出数轴上$n$个点，有$m$个操作，在时间$t$让一个点以一定的速度移动，或者询问时间$t$时距离原点最远的点。

Solution

　　超哥线段树。时间当做横坐标，负半轴的情况斜率和截距乘上$-1$再做一遍。

　　话说我为什么要离散化=  =

细节

　　LL

代码

// bzoj3938
#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<cmath>
#define LL long long
#define inf (1ll<<60)
#define eps 1e-8
#define free(a) freopen(a".in","r",stdin),freopen(a".out","w",stdout)
using namespace std;

const int maxn=1000010;
LL ans[maxn],a[maxn];
int n,m,cnt,tot,last[maxn],tim[maxn],q[maxn];
char ch[100];

struct seg {
	LL k,b,l,r;
	LL ch(LL x) {return k*x+b;}
}v[maxn];
struct node {int cov;seg mx;}tr[maxn<<2];
double X(seg a,seg b) {
	return a.k==b.k ? inf : 1.0*(a.b-b.b)/(b.k-a.k);
}

void insert(int k,int l,int r,int s,int t,seg g) {
	int mid=(l+r)>>1;
	if (l==s && r==t) {
		if (!tr[k].cov) tr[k].mx=g,tr[k].cov=1;
		else {
			seg a1=tr[k].mx,a2=g;double x=X(a1,a2);
			if (a1.ch(tim[l])<a2.ch(tim[l]) || (fabs(a1.ch(tim[l])-a2.ch(tim[l]))<eps && a1.k<a2.k)) swap(a1,a2);
			if (x<=tim[l] || x>=tim[r]) {tr[k].mx=a1;return;}
			if (x<=tim[mid]) tr[k].mx=a2,insert(k<<1,l,mid,s,mid,a1);
			else tr[k].mx=a1,insert(k<<1|1,mid+1,r,mid+1,t,a2);
		}
		return;
	}
	if (t<=mid) insert(k<<1,l,mid,s,t,g);
	else if (s>mid) insert(k<<1|1,mid+1,r,s,t,g);
	else insert(k<<1,l,mid,s,mid,g),insert(k<<1|1,mid+1,r,mid+1,t,g);
}
LL query(int k,int l,int r,int x) {
	if (l==r) return tr[k].cov ? tr[k].mx.ch(x) : 0;
	int mid=(l+r)>>1;LL t=0;
	if (x<=tim[mid]) t=query(k<<1,l,mid,x);
	else t=query(k<<1|1,mid+1,r,x);
	return tr[k].cov ? max(t,tr[k].mx.ch(x)) : t;
}
void clear(int k,int l,int r) {
	tr[k].cov=0;
	if (l==r) return;
	int mid=(l+r)>>1;
	clear(k<<1,l,mid);
	clear(k<<1|1,mid+1,r);
}

int main() {
	scanf("%d%d",&n,&m);
	for (int i=1;i<=n;i++) scanf("%lld",&a[i]);
	for (int i=1;i<=n;i++) v[++cnt]=(seg){0,a[i],0,-1},last[i]=cnt;
	for (int x,i=1;i<=m;i++) {
		scanf("%d",&tim[i]);
		scanf("%s",ch);
		if (ch[0]=='c') {
			scanf("%d%lld",&x,&v[++cnt].k);
			v[last[x]].r=v[cnt].l=tim[i];v[cnt].r=-1;
			a[x]=v[last[x]].k*tim[i]+v[last[x]].b;
			v[cnt].b=a[x]-v[cnt].k*tim[i];
			last[x]=cnt;
		}
		if (ch[0]=='q') q[++tot]=tim[i];
	}
	m=unique(tim,tim+1+m)-tim-1;
	for (int i=1;i<=cnt;i++) {
		if (v[i].r==-1) v[i].r=tim[m];
		int l=lower_bound(tim,tim+1+m,v[i].l)-tim;
		int r=lower_bound(tim,tim+1+m,v[i].r)-tim;
		insert(1,0,m,l,r,v[i]);
	}
	for (int i=1;i<=tot;i++)
		ans[i]=query(1,0,m,q[i]);
	clear(1,0,m);
	for (int i=1;i<=cnt;i++) {
		int l=lower_bound(tim,tim+1+m,v[i].l)-tim;
		int r=lower_bound(tim,tim+1+m,v[i].r)-tim;
		v[i].k*=-1;v[i].b*=-1;
		insert(1,0,m,l,r,v[i]);
	}
	for (int i=1;i<=tot;i++) ans[i]=max(ans[i],query(1,0,m,q[i]));
	for (int i=1;i<=tot;i++) printf("%lld\n",ans[i]);
	return 0;
}