分析：
每次操作把以前没有出现这个数的设为1，有这个数的设为0。首先将当前区间设为1，考虑有set维护这个颜色出现的区间，然后把所有与当前区间相交的拿出来，修改为0。
复杂度？每次操作的线段只会加入到一次set中，从set中取出一次，只会修改一次，然后就合并成大的了，每次操作也只会加入一条线段。
代码：
#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>
#include<cctype>
#include<set>
#include<queue>
#include<vector>
#include<map>
#include<bitset>
#define pa pair<int,int>
using namespace std;
typedef long long LL; inline int read() {
int x=,f=;char ch=getchar();for(;!isdigit(ch);ch=getchar())if(ch=='-')f=-;
for(;isdigit(ch);ch=getchar())x=x*+ch-'';return x*f;
} const int N = ;
set< pa > s[N];
int n; struct SegmentTree{
int sum[N << ], tag[N << ], n;
SegmentTree() { memset(tag, -, sizeof(tag)); }
void pushdown(int rt,int len) {
tag[rt << ] = tag[rt], tag[rt << | ] = tag[rt];
sum[rt << ] = tag[rt] * (len - len / );
sum[rt << | ] = tag[rt] * (len / );
tag[rt] = -;
}
void update(int l,int r,int rt,int L,int R,int v) {
if (L > R) return ;
if (L <= l && r <= R) { tag[rt] = v, sum[rt] = v * (r - l + ); return ; }
if (~tag[rt]) pushdown(rt, r - l + );
int mid = (l + r) >> ;
if (L <= mid) update(l, mid, rt << , L, R, v);
if (R > mid) update(mid + , r, rt << | , L, R, v);
sum[rt] = sum[rt << ] + sum[rt << | ];
}
int query(int l,int r,int rt,int L,int R) {
if (L <= l && r <= R) return sum[rt];
if (~tag[rt]) pushdown(rt, r - l + );
int mid = (l + r) >> , res = ;
if (L <= mid) res = query(l, mid, rt << , L, R);
if (R > mid) res += query(mid + , r, rt << | , L, R);
return res;
}
}T;
void Insert() {
int x = read(), y = read(), k = read(), l = max(, x - k), r = min(n, x + k), nowl = l, nowr = r;
if (s[y].empty()) {
T.update(, n, , l, r, );
s[y].insert(pa(l, r)); return ;
}
set< pa > :: iterator it = s[y].lower_bound(pa(l, ));
set< pa > :: iterator pre = it; if (it != s[y].begin()) {
pre --;
if (pre->second >= l) it = pre;
}
if (it == s[y].end() || (it->first) > r) {
T.update(, n, , l, r, );
s[y].insert(pa(l, r)); return ;
}
nowl = min(nowl, it->first);
T.update(, n, , l, r, );
while (it != s[y].end() && (it->first) <= r) {
T.update(, n, , max(l, it->first), min(r, it->second), );
nowr = max(nowr, it->second);
pre = it; it ++; s[y].erase(pre);
}
s[y].insert(pa(nowl, nowr));
}
void query() {
int l = read(), r = read();
printf("%d\n", T.query(, n, , l, r));
}
int main() {
n = read();int m = read();
for (int i = ; i <= m; ++i) {
if (read() == ) Insert();
else query();
}
return ;
}