题解联

2021-07-23 09:04:35 作者：互联网

这题考场上居然调出来了没炸细节……

第一思路是链表，然而链表上没法二分所以是\(n^2\)的
要不是数据范围很大就可以用线段树了
考虑离散化，但离散化完两个不连续的位置可能会被离散得连续了
所以把每个\(l-1, l, r, r+1\)全扔进去离散化，同时注意把1也扔进去就行了
我打的是全0/1和异或标记，注意这两个标记不共存即可

Code:

#include <bits/stdc++.h>
using namespace std;
#define INF 0x3f3f3f3f
#define N 100010
#define ll long long 
#define ld long double
#define usd unsigned
#define ull unsigned long long
#define int long long 

#define getchar() (p1==p2&&(p2=(p1=buf)+fread(buf, 1, 1<<21, stdin)), p1==p2?EOF:*p1++)
char buf[1<<21], *p1=buf, *p2=buf;
inline int read() {
	int ans=0, f=1; char c=getchar();
	while (!isdigit(c)) {if (c=='-') f=-f; c=getchar();}
	while (isdigit(c)) {ans=(ans<<3)+(ans<<1)+(c^48); c=getchar();}
	return ans*f;
}

int m;

namespace force{
	bool vis[N];
	void solve() {
		for (int i=1,t,l,r; i<=m; ++i) {
			t=read(); l=read(); r=read();
			if (t==1) {
				for (int j=l; j<=r; ++j) vis[j]=1;
			}
			else if (t==2) {
				for (int j=l; j<=r; ++j) vis[j]=0;
			}
			else {
				for (int j=l; j<=r; ++j) vis[j]^=1;
			}
			for (int j=1; j<N; ++j) if (!vis[j]) {printf("%lld\n", j); break;}
		}
		exit(0);
	}
}

namespace task1{
	int uni[N*10], usize;
	int qt[N], ql[N], qr[N];
	const int SIZE=(N*8)<<2;
	int tl[SIZE], tr[SIZE];
	bool tag0[SIZE], tag1[SIZE], tagx[SIZE];
	#define tl(p) tl[p]
	#define tr(p) tr[p]
	#define tag0(p) tag0[p]
	#define tag1(p) tag1[p]
	#define tagx(p) tagx[p]
	void build(int p, int l, int r) {
		tl(p)=l; tr(p)=r; tag0(p)=1;
		if (l>=r) return ;
		int mid=(l+r)>>1;
		build(p<<1, l, mid);
		build(p<<1|1, mid+1, r);
	}
	void spread(int p) {
		if (tag0(p) || tag1(p)) {
			tag0(p<<1)=tag0(p), tag1(p<<1)=tag1(p), tagx(p<<1)=0;
			tag0(p<<1|1)=tag0(p), tag1(p<<1|1)=tag1(p), tagx(p<<1|1)=0;
			//assert(!tagx(p));
			return ;
		}
		if (!tagx(p)) return ;
		
		//assert(!(tag0(p)&&tag1(p)));
		if (tag0(p<<1)||tag1(p<<1)) {swap(tag0(p<<1), tag1(p<<1)); tagx(p<<1)=0;}
		else tagx(p<<1)^=1;
		
		if (tag0(p<<1|1)||tag1(p<<1|1)) {swap(tag0(p<<1|1), tag1(p<<1|1)); tagx(p<<1|1)=0;}
		else tagx(p<<1|1)^=1;
		tagx(p)=0;
	}
	void upd(int p, int ql, int qr, int dat) {
		//cout<<"upd: "<<p<<' '<<tl(p)<<' '<<tr(p)<<' '<<ql<<' '<<qr<<' '<<dat<<endl;
		if (ql<=tl(p)&&qr>=tr(p)) {
			if (dat==0) {tag0(p)=tag1(p)=tagx(p)=0; tag0(p)=1; return ;}
			if (dat==1) {tag0(p)=tag1(p)=tagx(p)=0; tag1(p)=1; return ;}
			if (dat==2) {
				//cout<<"x1: "<<p<<' '<<tl(p)<<' '<<tr(p)<<' '<<tag0(p)<<' '<<tag1(p)<<' '<<tagx(p)<<endl;
				if (tag0(p)||tag1(p)) swap(tag0(p), tag1(p)), tagx(p)=0; //, cout<<"case1"<<endl;
				else tagx(p)^=1; //, cout<<"case2"<<endl;
				//cout<<"x2: "<<p<<' '<<tl(p)<<' '<<tr(p)<<' '<<tag0(p)<<' '<<tag1(p)<<' '<<tagx(p)<<endl;
				return ;
			}
		}
		spread(p);
		tag0(p)=tag1(p)=0;
		//cout<<"zero "<<p<<endl;
		int mid=(tl(p)+tr(p))>>1;
		if (ql<=mid) upd(p<<1, ql, qr, dat);
		if (qr>mid) upd(p<<1|1, ql, qr, dat);
	}
	int query(int p) {
		//cout<<"query "<<p<<' '<<tl(p)<<' '<<tr(p)<<endl;
		if (!p) return -1;
		//assert(!(tag0(p)&&tag1(p)));
		if (tag0(p)) return tl(p);
		if (tag1(p)) return -1;
		if (tl(p)==tr(p)) return -1;
		spread(p);
		int t, ans=-1;
		t=query(p<<1);
		if (t!=-1) {
			if (ans==-1) ans=t;
			else ans=min(ans, t);
		}
		if (ans!=-1) return ans;
		t=query(p<<1|1);
		if (t!=-1) {
			if (ans==-1) ans=t;
			else ans=min(ans, t);
		}
		return ans;
	}
	void solve() {
		//cout<<double(sizeof(tl)*4+sizeof(uni))/1024/1024<<endl;
		for (int i=1; i<=m; ++i) {
			qt[i]=read(); ql[i]=read(); qr[i]=read();
			uni[++usize]=ql[i]; if (ql[i]-1>0) uni[++usize]=ql[i]-1; uni[++usize]=ql[i]+1;
			uni[++usize]=qr[i]; if (qr[i]-1>0) uni[++usize]=qr[i]-1; uni[++usize]=qr[i]+1;
		}
		uni[++usize]=1;
		sort(uni+1, uni+usize+1);
		usize=unique(uni+1, uni+usize+1)-uni-1;
		//cout<<"usize: "<<usize<<endl;
		//cout<<"uni: "; for (int i=1; i<=usize; ++i) cout<<uni[i]<<' '; cout<<endl;
		build(1, 1, usize+1);
		int tsk[]={-1, 1, 0, 2};
		for (int i=1,t; i<=m; ++i) {
			upd(1, lower_bound(uni+1, uni+usize+1, ql[i])-uni, lower_bound(uni+1, uni+usize+1, qr[i])-uni, tsk[qt[i]]);
			//cout<<"upd: "<<lower_bound(uni+1, uni+usize+1, ql[i])-uni<<' '<<lower_bound(uni+1, uni+usize+1, qr[i])-uni<<' '<<tsk[qt[i]]<<endl;
			t = query(1);
			//cout<<"t: "<<t<<endl;
			printf("%lld\n", uni[t]);
		}
		exit(0);
	}
}

signed main()
{
	#ifdef DEBUG
	freopen("1.in", "r", stdin);
	#endif
	
	m=read();
	//force::solve();
	task1::solve();

	return 0;
}

标签：qr,++,题解,long,usize,uni,define
来源： https://www.cnblogs.com/narration/p/15047308.html

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