题意
维护一个\(01\)串，一开始全部都是\(0\)
\(3\)种操作
\(1.\)把一个区间都变为\(1\)
\(2.\)把一个区间都变为\(0\)
\(3.\)把一个区间的所有数字翻转过来
每次操作完成之后询问区间最小的\(0\)的位置
\(l,r<=10^{18}\)
Sol
直接上线段树，如果这个区间左边是满的就去右边，否则去左边
动态开点，加上\(lazy\)
但这样会爆空间
所以把区间离散化
注意要离散\(1\)，\(l\)，\(r\)，\(l+1\)，\(r+1\)
这些都可能会出现答案
# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(4e5 + 5);
IL ll Input(){
RG ll x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}
int n, len, opt[_];
ll ql[_], qr[_], o[_];
struct Segment{
int sum, rev, tag;
} T[_ << 2];
IL void Adjust1(RG int x, RG int l, RG int r, RG int v){
T[x].rev = 0, T[x].tag = v--;
T[x].sum = v * (r - l + 1);
}
IL void Adjust2(RG int x, RG int l, RG int r){
T[x].sum = r - l + 1 - T[x].sum;
T[x].rev ^= 1;
}
IL void Pushdown(RG int x, RG int l, RG int r){
RG int mid = (l + r) >> 1, ls = x << 1, rs = x << 1 | 1;
if(T[x].tag){
Adjust1(ls, l, mid, T[x].tag);
Adjust1(rs, mid + 1, r, T[x].tag);
T[x].tag = 0;
}
if(T[x].rev){
Adjust2(ls, l, mid);
Adjust2(rs, mid + 1, r);
T[x].rev = 0;
}
}
IL void Modify(RG int x, RG int l, RG int r, RG int L, RG int R, RG int v){
if(L <= l && R >= r){
Adjust1(x, l, r, v);
return;
}
Pushdown(x, l, r);
RG int mid = (l + r) >> 1;
if(L <= mid) Modify(x << 1, l, mid, L, R, v);
if(R > mid) Modify(x << 1 | 1, mid + 1, r, L, R, v);
T[x].sum = T[x << 1].sum + T[x << 1 | 1].sum;
}
IL void Reverse(RG int x, RG int l, RG int r, RG int L, RG int R){
if(L <= l && R >= r){
Adjust2(x, l, r);
return;
}
Pushdown(x, l, r);
RG int mid = (l + r) >> 1;
if(L <= mid) Reverse(x << 1, l, mid, L, R);
if(R > mid) Reverse(x << 1 | 1, mid + 1, r, L, R);
T[x].sum = T[x << 1].sum + T[x << 1 | 1].sum;
}
IL int Query(RG int x, RG int l, RG int r){
if(l == r) return l;
Pushdown(x, l, r);
RG int mid = (l + r) >> 1, ans;
if(T[x << 1].sum != mid - l + 1) ans = Query(x << 1, l, mid);
else ans = Query(x << 1 | 1, mid + 1, r);
T[x].sum = T[x << 1].sum + T[x << 1 | 1].sum;
return ans;
}
int main(RG int argc, RG char *argv[]){
n = Input(), o[++len] = 1;
for(RG int i = 1; i <= n; ++i){
opt[i] = Input(), ql[i] = Input(), qr[i] = Input();
o[++len] = ql[i], o[++len] = qr[i];
o[++len] = ql[i] + 1, o[++len] = qr[i] + 1;
}
sort(o + 1, o + len + 1), len = unique(o + 1, o + len + 1) - o - 1;
for(RG int i = 1; i <= n; ++i){
ql[i] = lower_bound(o + 1, o + len + 1, ql[i]) - o;
qr[i] = lower_bound(o + 1, o + len + 1, qr[i]) - o;
if(opt[i] == 1) Modify(1, 1, len, ql[i], qr[i], 2);
else if(opt[i] == 2) Modify(1, 1, len, ql[i], qr[i], 1);
else Reverse(1, 1, len, ql[i], qr[i]);
printf("%lld\n", o[Query(1, 1, len)]);
}
return 0;
}