题意
Sol
直接拿set维护\(li\)连续段。因为set内的区间互不相交，而且每个线段会被至多加入删除一次，所以复杂度是对的。
#include<bits/stdc++.h>
#define Pair pair<int, int>
#define MP make_pair
#define fi first
#define se second
using namespace std;
const int MAXN = 1e6 + 10, INF = 2147483646;
inline int read() {
char c = getchar(); int x = 0, f = 1;
while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
return x * f;
}
int N, M, s[MAXN], f[MAXN], ll[MAXN], rr[MAXN];
#define ls k << 1
#define rs k << 1 | 1
void update(int k) {
s[k] = s[ls] + s[rs];
}
void ps(int k, int v) {
if(v == 1) s[k] = rr[k] - ll[k] + 1;
else s[k] = 0;
f[k] = v;
}
void pushdown(int k) {
if(f[k] == -1) return ;
ps(ls, f[k]); ps(rs, f[k]);
f[k] = -1;
}
void Build(int k, int l, int r) {
ll[k] = l; rr[k] = r;
if(l == r) return ;
int mid = l + r >> 1;
Build(ls, l, mid);
Build(rs, mid + 1, r);
}
void Mem(int k, int l, int r, int ql, int qr, int val) {
if(ql > qr) return ;
if(ql <= l && r <= qr) {ps(k, val); return ;}
int mid = l + r >> 1;
pushdown(k);
if(ql <= mid) Mem(ls, l, mid, ql, qr, val);
if(qr > mid) Mem(rs, mid + 1, r, ql, qr, val);
update(k);
}
int Query(int k, int l, int r, int ql, int qr) {
if(ql > qr) return 0;
int ans = 0;
//cout << ans << '\n';
if(ql <= l && r <= qr) return s[k];
int mid = l + r >> 1;
pushdown(k);
if(ql <= mid) ans += Query(ls, l, mid, ql, qr);
if(qr > mid) ans += Query(rs, mid + 1, r, ql, qr);
return ans;
}
set<Pair> S[MAXN];
#define sit set<Pair>::iterator
void Add(int l, int r, int x) {
int pl = l, pr = r;
set<Pair> &s = S[x];
sit it = s.lower_bound(MP(l, r));
Mem(1, 1, N, l, r, 1);
if(it != s.begin()) {
it--;
if(it -> se > r) {
Mem(1, 1, N, max(pl, it -> fi), min(pr, it -> se), 0);
}
if(it -> se >= l ) {
l = min(l, it -> fi); r = max(r, it -> se);
Mem(1, 1, N, max(pl, it -> fi), min(pr, it -> se), 0);
s.erase(it++);
}
}
it = s.lower_bound(MP(l, r));
while((it -> se >= l && it -> se <= r) || (it -> fi >= l && it -> fi <= r)) {
l = min(l, it -> fi); r = max(r, it -> se);
Mem(1, 1, N, max(pl, it -> fi), min(pr, it -> se), 0);
s.erase(it++);
}
s.insert(MP(l, r));
}
signed main() {
// freopen("a.in", "r", stdin);
N = read(); M = read();
Build(1, 1, N);
for(int i = 1; i <= N; i++) S[i].insert(MP(INF, INF));
while(M--) {
int opt = read();
if(opt == 1) {
int xi = read(), val = read(), ki = read();
int l = max(1, xi - ki), r = min(N, xi + ki);
Add(l, r, val);
}
else {
int l = read(), r = read();
printf("%d\n", Query(1, 1, N, l, r));
}
}
return 0;
}