CodeForces 915E Physical Education Lessons 線段樹動態開點

題目鏈接：https://codeforces.com/problemset/problem/915/E
題意：q次操作，每次將一段區間置0或者置1，然後輸出1~n中0的個數
思路：由於n太大，所以我們考慮線段樹動態開點，可以大大減少點的數量便於維護，這也是一道動態開點的模板題

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
#define fi first
#define se second
#define ls rt << 1
#define rs rt << 1|1
#define po pop_back
#define pb push_back
#define mk make_pair
#define lson l, mid, ls
#define rson mid + 1, r, rs
#define pll pair<ll, ll>
#define pii pair<int, int>
#define ull unsigned long long
#define pdd pair<double, double>
const int mod = 1e9 + 7;
const int maxn = 3e5 + 10;
const int inf = 0x3f3f3f3f;
const ll linf = 0x3f3f3f3f3f3f3f3f;
struct seg
{
    int lc, rc, val, lazy;
}tree[maxn * 50];
int root = 0, tot = 0; //根結點和結點編號，多樣例初始化

int build()
{
    ++tot;
    tree[tot].lc = tree[tot].rc = tree[tot].val = 0;
    tree[tot].lazy = -1; //
    return tot;
}
void push_up(int rt)
{
    tree[rt].val = tree[tree[rt].lc].val + tree[tree[rt].rc].val;
}
void push_down(int l, int r, int rt)
{
    if(tree[rt].lazy != -1 && l != r)
    {
        int mid = (l + r) >> 1;
        if(!tree[rt].lc) tree[rt].lc = build();
        if(!tree[rt].rc) tree[rt].rc = build();
        tree[tree[rt].lc].val = tree[rt].lazy * (mid - l + 1);
        tree[tree[rt].lc].lazy = tree[rt].lazy;
        tree[tree[rt].rc].val = tree[rt].lazy * (r - mid);
        tree[tree[rt].rc].lazy = tree[rt].lazy;
        tree[rt].lazy = -1;
    }
}
void update(int ql, int qr, int val, int l, int r, int &rt) //區間修改 傳遞引用
{
    if(!rt) rt = build();
    if(ql <= l && r <= qr)
    {
        tree[rt].val = val * (r - l + 1);
        tree[rt].lazy = val;
        return;
    }
    int mid = (l + r) >> 1;
    push_down(l, r, rt);
    if(ql <= mid) update(ql, qr, val, l, mid, tree[rt].lc);
    if(qr > mid) update(ql, qr, val, mid + 1, r, tree[rt].rc);
    push_up(rt);
}
int query(int ql, int qr, int l, int r, int rt)
{
    if(!rt) return 0;
    if(ql <= l && qr <= r)
        return tree[rt].val;
    int mid = (l + r) >> 1, res = 0;
    if(ql <= mid)
        res += query(ql, qr, l, mid, tree[rt].lc);
    if(qr > mid)
        res += query(ql, qr, mid + 1, r, tree[rt].rc);
    return res;
}

int main()
{
    int n, q;
    scanf("%d%d", &n, &q);
    while(q--)
    {
        int l, r, k;
        scanf("%d%d%d", &l, &r, &k);
        update(l, r, k % 2, 1, n, root);
        printf("%d\n", n - query(1, n, 1, n, root));
    }
	return 0;
}