set维护连续段+线段树：1018T2

本文搬运自本人高中时期CSDN博客，若图片加载不出来，可到原文查看：https://blog.csdn.net/zhangtingxiqwq/article/details/133914164

http://cplusoj.com/d/senior/p/386?tid=652f5fe6c1fe41bc229c18fb

线段树维护01，和，支持翻转操作

用类似珂朵莉树的方法维护连续段，连续段之间分别统计，取max

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#include<bits/stdc++.h>
using namespace std;
#define int long long
inline int read(){int x=0,f=1;char ch=getchar(); while(ch<'0'||
ch>'9'){if(ch=='-')f=-1;ch=getchar();}while(ch>='0'&&ch<='9'){
x=(x<<1)+(x<<3)+(ch^48);ch=getchar();}return x*f;}
#define Z(x) (x)*(x)
#define pb push_back
//mt19937 rand(time(0));
//mt19937_64 rand(time(0));
//srand(time(0));
#define N 1500010
//#define M
//#define mo
struct Ment {
	int l, r; 
	bool operator < (const Ment &A) const {
		return l < A.l; 
	}
};
struct ANS {
	int x, y; 
	void operator += (const ANS &A) {
		x += max(A.x, A.y); 
		y += min(A.x, A.y); 
	}
	void operator -= (const ANS &A) {
		x -= max(A.x, A.y); 
		y -= min(A.x, A.y); 
	}
	ANS operator + (const ANS &A) const {
		ANS B; 
		B.x = x + A.x; 
		B.y = y + A.y; 
		return B; 
	}
}ans;
struct QRY {
	int l, r, d; 
}qry[N];
int n, m, i, j, k, T;
int l, r, len, mn, mx, rt, v[N]; 
set<Ment>s; 
set<Ment>::iterator it, it1, it2; 

struct Segment_tree {
	int tot, ls[N<<2], rs[N<<2]; 
	int s[N<<2][2], c[N<<2][2], tag[N<<2]; 
	void push_up(int k) {
		s[k][0]=s[ls[k]][0]+s[rs[k]][0]; 
		s[k][1]=s[ls[k]][1]+s[rs[k]][1]; 
		c[k][0]=c[ls[k]][0]+c[rs[k]][0]; 
		c[k][1]=c[ls[k]][1]+c[rs[k]][1]; 
	}
	void print(int k, int l, int r) {
		printf("# %d : [%d %d]\n", k, l, r); 
		printf("s : %lld %lld\n", s[k][0], s[k][1]); 
		printf("c : %lld %lld\n", c[k][0], c[k][1]); 
	}
	void build(int &k, int l, int r) {
		if(!k) k=++tot; 
		if(l==r) {
			int len=v[l+1]-v[l]; // [v(l), v(l+1)-1]
			if(v[l+1]%2==0) c[k][0]=(len+1)/2, c[k][1]=len/2; 
			else c[k][1]=(len+1)/2, c[k][0]=len/2; 
			return ; 
		}
		int mid = (l+r)>>1; 
		build(ls[k], l, mid); 
		build(rs[k], mid+1, r); 
		push_up(k); 
//		print(k, l, r); 
	}
	void modify(int k, int h) { //k节点加上h 
		tag[k]+=h; 
		s[k][0]+=(c[k][0]+c[k][1])*(h/2); 
		s[k][1]+=(c[k][0]+c[k][1])*(h/2); 
		if(h % 2 == 0) return ; 
		swap(c[k][0], c[k][1]); 
		s[k][0]+=c[k][0]; s[k][1]+=c[k][1]; 
	}
	void push_down(int k) {
		modify(ls[k], tag[k]); 
		modify(rs[k], tag[k]); 
		tag[k]=0; 
	}
	void add(int k, int l, int r, int x, int y, int h) {
		if(l>=x && r<=y) return modify(k, h), void(); 		
		int mid = (l+r)>>1; push_down(k); 
		if(x <= mid) add(ls[k], l, mid, x, y, h); 
		if(y >= mid+1) add(rs[k], mid+1, r, x, y, h); 
		push_up(k); 
	}
	ANS que(int k, int l, int r, int x, int y) {
		if(l>=x && r<=y) return {s[k][0], s[k][1]}; 
		int mid = (l+r)>>1; push_down(k); 
		ANS A, B; A.x=A.y=B.x=B.y=0; 
		if(x <= mid) A = que(ls[k], l, mid, x, y); 
		if(y >= mid+1) B = que(rs[k], mid+1, r, x, y); 
		push_up(k); 
		return A + B; 
	}
}Seg;

signed main()
{
//	freopen("in.txt", "r", stdin);
//	freopen("out.txt", "w", stdout);
	freopen("nvwa.in", "r", stdin);
	freopen("nvwa.out", "w", stdout);
//	T=read();
//	while(T--) {
//
//	}
	auto Add = [&] (int x) -> void {
		v[++j]=x; 
		if(x-2>=1) v[++j]=x-2; 
		if(x-1>=1) v[++j]=x-1; 
		if(x+1<=n) v[++j]=x+1; 
		if(x+2<=n) v[++j]=x+2; 
//		if(x+3<=n) v[++j]=x+3; 
	}; 
	n=read()+10; T=read(); 
	v[++j]=1; v[++j]=n; 
	for(i=1; i<=T; ++i) {
		qry[i].l=read()+1+1+1; qry[i].r=read()+1+1; qry[i].d=read(); 
		Add(qry[i].l); Add(qry[i].r); 
	}
	sort(v+1, v+j+1); 
	len = unique(v+1, v+j+1) - v - 1; v[len+1]=n+1; 
//	for(i=1; i<=len+1; ++i) printf("%lld ", v[i]); printf("\n"); 
	Seg.build(rt, 1, len); 
	for(i=1; i<=len; ++i) s.insert({i, i}); 
	for(i=1; i<=T; ++i) {
		l = qry[i].l; r = qry[i].r; 
		l = lower_bound(v+1, v+len+1, l) - v; 
		r = lower_bound(v+1, v+len+1, r) - v; 
// 		printf(">> %lld %lld\n", l, r); 
		it1 = s.lower_bound({l, 0}); --it1; 
		it2 = s.lower_bound({r+1, 0}); 
		mn = 1e9; mx = -1e9; 
//		printf("> %lld %lld\n", ans.y, ans.x); 
		for(it = it1; it != it2; ++it) {
			mn = min(mn, (it->l)); 
			mx = max(mx, (it->r)); 
			ans -= Seg.que(1, 1, len, (it->l), (it->r)); 
//			printf("Del : [%lld %lld] => %lld %lld\n", (it->l), (it->r), ans.y, ans.x); 
		}
		s.erase(it1, it2); s.insert({mn, mx}); 
////		printf(">> [%lld %lld]\n", mn, mx); 
		Seg.add(1, 1, len, l, r, qry[i].d); 
		ans += Seg.que(1, 1, len, mn, mx); 
		printf("%lld %lld\n", ans.y, ans.x); 
////		for(auto t : s) printf("[%lld %lld] ", t.l, t.r); 
//		printf("\n"); 
	}
	return 0;	
}