链接：

https://codeforces.com/problemset/problem/915/E

题意：

This year Alex has finished school, and now he is a first-year student of Berland State University. For him it was a total surprise that even though he studies programming, he still has to attend physical education lessons. The end of the term is very soon, but, unfortunately, Alex still hasn‘t attended a single lesson!

Since Alex doesn‘t want to get expelled, he wants to know the number of working days left until the end of the term, so he can attend physical education lessons during these days. But in BSU calculating the number of working days is a complicated matter:

There are n days left before the end of the term (numbered from 1 to n), and initially all of them are working days. Then the university staff sequentially publishes q orders, one after another. Each order is characterised by three numbers l, r and k:

If k?=?1, then all days from l to r (inclusive) become non-working days. If some of these days are made working days by some previous order, then these days still become non-working days;
If k?=?2, then all days from l to r (inclusive) become working days. If some of these days are made non-working days by some previous order, then these days still become working days.
Help Alex to determine the number of working days left after each order!

思路：

求1-n里的0的个数。
动态开点线段树，对于每次更新，每个节点先判断是不是新的，是新的就重新建立即可。

代码：

#include<bits/stdc++.h>
using namespace std;
typedef long long LL;
const int MAXN = 4e5+10;
const int MOD = 1e9+7;
const int INF = 1e9;

int n, q;
int add[MAXN*40], sum[MAXN*40], ln[MAXN*40], rn[MAXN*40];
int rot = 0, cnt = 0;

void pushdown(int root, int l, int r)
{
    if (add[root] != -1)
    {
        if (!ln[root]) ln[root] = ++cnt;
        if (!rn[root]) rn[root] = ++cnt;
        int mid = (l+r)/2;
        add[ln[root]] = add[root];
        add[rn[root]] = add[root];
        sum[ln[root]] = add[root]*(mid-l+1);
        sum[rn[root]] = add[root]*(r-mid);
        add[root] = -1;
    }
}

void update(int &root, int l, int r, int ql, int qr, int v)
{
    if (!root) root = ++cnt;
    if (ql <= l && r <= qr)
    {
        add[root] = v;
        sum[root] = v*(r-l+1);
        return;
    }
    pushdown(root, l, r);
    int mid = (l+r)/2;
    if (ql <= mid) update(ln[root], l, mid, ql, qr, v);
    if (qr > mid) update(rn[root], mid+1, r, ql, qr, v);
    sum[root] = sum[ln[root]]+sum[rn[root]];
}

int main()
{
    //dsu on tree 启发式合并
    ios::sync_with_stdio(stdin);
    cin.tie(0), cout.tie(0);
    scanf("%d%d", &n, &q);
    memset(add, -1, sizeof(add));
    int l, r, k;
    while(q--)
    {
        scanf("%d%d%d", &l, &r, &k);
        update(rot, 1, INF, l, r, k == 2 ? 0 : 1);
        printf("%d\n", n-sum[1]);
    }

    return 0;
}

原文地址：https://www.cnblogs.com/YDDDD/p/12238018.html