思路：
这题必有平推操作，所以用珂朵莉写，极其方便，也不用考虑数据是否随机，根本卡不了，只需要平推后光速降复杂度即可，然后对贡献，只需暴力加减贡献即可。
参考代码：
/*
* @Author: vain
* @Date: 2020
* @LastEditTime: 2020-09-23 19:00:14
* @LastEditors: sueRimn
* @Description: 学不会 dp 的 fw
* @FilePath: \main\demo.cpp
*/
//#include <bits/stdc++.h>
#include <cstdio>
#include <algorithm>
#include <iostream>
#include <vector>
#include <map>
#include <queue>
#include <set>
#include <ctime>
#include <cstring>
#include <cstdlib>
#include <math.h>
#include <bitset>
using namespace std;
typedef long long ll;
#define ll long long
//typedef unsigned long long uint;
const int N = 1e3 + 20;
const int maxn = 1e5 + 20;
int a[maxn];
// typedef pair<int, int> p;
// priority_queue<p, vector<p>, greater<p>> m;
//int sum[maxn];
int max(int a, int b) {
return a > b ? a : b; }
int min(int a, int b) {
return a < b ? a : b; }
int gcd(int a, int b) {
return b ? gcd(b, a % b) : a; }
int lcm(int a, int b) {
return a * b / gcd(a, b); }
void swap(int &x, int &y) {
x ^= y, y ^= x, x ^= y; }
int lowbit(int x) {
return (x) & (-x); }
struct node
{
int l, r;
mutable int w;
node(int L = 0, int R = -1, int W = 0) : l(L), r(R), w(W){
};
bool operator<(const node &x) const {
return l < x.l; }
};
#define It set<node>::iterator
set<node> S;
int ans;
It split(int pos)
{
It it = S.lower_bound(pos);
if (it != S.end() && it->l == pos)
return it;
it--;
int L = it->l, R = it->r, w = it->w;
S.erase(it);
S.insert(node(L, pos - 1, w));
return S.insert(node(pos, R, w)).first;
}
void push(int l, int r, int w)
{
It itr = split(r + 1), itl = split(l);
int sum, res;
sum = res = 0;
for (It it = itl; it != itr; it++)
{
if ((it->w) && (w))
sum += it -> r - it->l + 1;
if (!(it->w) && !(w))
res += it -> r - it->l + 1;
}
S.erase(itl, itr);
S.insert(node(l, r, w));
if (w)
ans += r - l + 1 - sum;
else
ans += res - (r - l + 1);
}
int main()
{
// ios::sync_with_stdio(false);
// cin.tie(0), cout.tie(0);
int n, q;
scanf("%d %d", &n, &q);
S.insert(node(1, n, 1));
ans = n;
while (q--)
{
int k, l, r;
scanf("%d %d %d", &l, &r, &k);
push(l, r, k - 1);
printf("%d\n", ans);
}
}