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('
')
#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();
}