【LOJ】#3033. 「JOISC 2019 Day2」两个天线

LOJ#3033. 「JOISC 2019 Day2」两个天线

用后面的天线更新前面的天线，线段树上存历史版本的最大值

也就是线段树需要维护历史版本的最大值，后面的天线的标记中最大的那个和最小的那个，区间中最小的可用天线值，区间中最大的可用天线值

\(i\)可以被\(j\)用到，那么\(j\)在\([i + A_{i},i + B_{i}]\)中，我们枚举右端点的时候，假如到了\(i + A_{i}\)就把\(i\)标记为可用，如果到了\(i + B_{i} + 1\)就把\(i\)标记为不可用

然后枚举右端点，对于一个新加的端点，现在线段树中区间\([i - B_{i},i- A_{i}]\)是可用的，所以我们需要给这个区间打上标记

然后对于一个询问到了对应的右端点，只需要查找区间中历史版本的最大值就好了

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('\n')
#define eps 1e-10
#define MAXN 200005
#define ba 47
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int l,r,mn,mx,lzmn,lzmx,oldmx;
}tr[MAXN * 4];
int N,Q,H[MAXN],A[MAXN],B[MAXN],ans[MAXN];
int ql[MAXN],qr[MAXN];
vector<int> ed[MAXN],st[MAXN],qe[MAXN];
void update(int u) {
    tr[u].mn = min(tr[u << 1].mn,tr[u << 1 | 1].mn);
    tr[u].mx = max(tr[u << 1].mx,tr[u << 1 | 1].mx);
    tr[u].oldmx = max(tr[u << 1].oldmx,tr[u << 1 | 1].oldmx);
}
void addlz(int u,int v) {
    tr[u].oldmx = max(tr[u].oldmx,v - tr[u].mn);
    tr[u].oldmx = max(tr[u].oldmx,tr[u].mx - v);
    tr[u].lzmn = min(v,tr[u].lzmn);
    tr[u].lzmx = max(v,tr[u].lzmx);
}
void pushdown(int u) {
    if(tr[u].lzmn <= 1e9) {
	addlz(u << 1,tr[u].lzmn);
	addlz(u << 1 | 1,tr[u].lzmn);
	tr[u].lzmn = 2e9;
    }
    if(tr[u].lzmx > 0) {
	addlz(u << 1,tr[u].lzmx);
	addlz(u << 1 | 1,tr[u].lzmx);
	tr[u].lzmx = 0;
    }
}
void build(int u,int l,int r) {
    tr[u].l = l;tr[u].r = r;
    tr[u].mn = 1e9 + 1,tr[u].mx = 0;
    tr[u].lzmn = 2e9;tr[u].lzmx = 0;tr[u].oldmx = -1;
    if(l == r) return;
    int mid = (l + r) >> 1;
    build(u << 1,l,mid);
    build(u << 1 | 1,mid + 1,r);    
}
void change(int u,int pos,int op) {
    if(tr[u].l == tr[u].r) {
	if(op == 1) tr[u].mn = tr[u].mx = H[pos];
	else {
	    tr[u].mn = 1e9 + 1,tr[u].mx = 0;
	}
	return;
    }
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(pos <= mid) change(u << 1,pos,op);
    else if(pos > mid) change(u << 1 | 1,pos,op);
    update(u);
}
void add(int u,int l,int r,int v) {
    if(tr[u].l == l && tr[u].r == r) {
	addlz(u,v);return;
    }
    pushdown(u);
    int mid = (tr[u].l + tr[u].r) >> 1;
    if(r <= mid) add(u << 1,l,r,v);
    else if(l > mid) add(u << 1 | 1,l,r,v);
    else {add(u << 1,l,mid,v);add(u << 1 | 1,mid + 1,r,v);}
    update(u);
}
int Query(int u,int l,int r) {
    if(tr[u].l == l && tr[u].r == r) return tr[u].oldmx;
    int mid = (tr[u].l + tr[u].r) >> 1;
    pushdown(u);
    if(r <= mid) return Query(u << 1,l,r);
    else if(l > mid) return Query(u << 1 | 1,l,r);
    else {return max(Query(u << 1,l,mid),Query(u << 1 | 1,mid + 1,r));}
}
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) {
	read(H[i]);read(A[i]);read(B[i]);
	int l = i + A[i],r = min(i + B[i],N);
	if(l <= r) {st[l].pb(i),ed[r + 1].pb(i);}
    }
    read(Q);
    for(int i = 1 ; i <= Q ; ++i) {
	read(ql[i]);read(qr[i]);
	qe[qr[i]].pb(i);
    }
    build(1,1,N);
    for(int i = 1 ; i <= N ; ++i) {
	for(auto t : st[i]) change(1,t,1);
	for(auto t : ed[i]) change(1,t,-1);
	int l = i - B[i],r = i - A[i];
	l = max(l,1);
	if(l <= r) add(1,l,r,H[i]);
	for(auto id : qe[i]) {
	    ans[id] = Query(1,ql[id],qr[id]);
	}
    }
    for(int i = 1 ; i <= Q ; ++i) {
	out(ans[i]);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}