noip2017集训测试赛 Problem B: mex [补档]

Posted 2020-10-22 Zeonfai

Description

给你一个无限长的数组，初始的时候都为0，有3种操作：
操作1是把给定区间[l,r][l,r] 设为1，
操作2是把给定区间[l,r][l,r] 设为0，
操作3把给定区间[l,r][l,r] 0,1反转。
一共n个操作，每次操作后要输出最小位置的0。

Input

第一行一个整数n，表示有n个操作
接下来n行，每行3个整数op,l,r表示一个操作

Output

共n行，一行一个整数表示答案

Sample Input

3
1 3 4
3 1 6
2 1 3

Sample Output

1
3
1

HINT

对于30%的数据\(1≤n≤10^3,1≤l≤r≤10^{18}1≤n≤10^3,1≤l≤r≤10^{18}\)
对于100%的数据\(1≤n≤10^5,1≤l≤r≤10^{18}1≤n≤10^5,1≤l≤r≤10^{18}\)

Solution

正解时空复杂度为\(O(n \log n)\).
我的解法时空复杂度也是\(O(n \log n)\)
但是它就是MLE了那么一点.

正解: 离散化 + 线段树.
我的做法: 开两棵离散线段树, 分别代表0和1, 每次把一棵中的一些点拆下来放到另一棵里面.

/*
mind the value of INF
*/

#include <cstdio>
#include <cctype>

namespace Zeonfai
{
    inline long long getInt()
    {
        long long a = 0, sgn = 1; char c;
        while(! isdigit(c = getchar())) if(c == ‘-‘) sgn *= -1;
        while(isdigit(c)) a = a * 10 + c - ‘0‘, c = getchar();
        return a * sgn;
    }
}
const long long INF = (long long)1e18 + 7;
// const long long INF = 11;
struct segmentTree
{
    struct node
    {
        node *suc[2];
        long long L, R;
        long long vst, sz;
        inline node(long long _L, long long _R)
        {
            L = _L; R = _R; vst = 0; sz = R - L + 1;
            for(long long i = 0; i < 2; ++ i) suc[i] = NULL;
        }
    }*rt;
    struct result
    {
        node *a, *b;
        inline result(node *_a, node *_b) {a = _a; b = _b;}
    };
    inline void initialize() {rt = new node(1, INF);}
    result split(node *u, long long L, long long R)
    {
        if(u == NULL || (u->L >= L && u->R <= R)) return result(u, NULL);
        long long mid = u->L + u->R >> 1;
        if(! u->vst)
        {
            u->suc[0] = new node(u->L, mid); u->suc[1] = new node(mid + 1, u->R);
            u->vst = 1;
        }
        node *_u = new node(u->L, u->R); _u->vst = 1;
        if(L <= mid)
        {
            result res = split(u->suc[0], L, R);
            _u->suc[0] = res.a; u->suc[0] = res.b;
        }
        if(R > mid)
        {
            result res = split(u->suc[1], L, R);
            _u->suc[1] = res.a; u->suc[1] = res.b;
        }
        u->sz = 0;
        for(long long i = 0; i < 2; ++ i) if(u->suc[i] != NULL) u->sz += u->suc[i]->sz;
        _u->sz = 0;
        for(long long i = 0; i < 2; ++ i) if(_u->suc[i] != NULL) _u->sz += _u->suc[i]->sz;
        return result(_u, u);
    }
    inline node* split(long long L, long long R)
    {
        result res = split(rt, L, R);
        rt = res.b; return res.a;
    }
    inline node* merge(node *u, node *_u)
    {
        if(u == NULL) return _u; if(_u == NULL) return u;
        for(long long i = 0; i < 2; ++ i) u->suc[i] = merge(u->suc[i], _u->suc[i]);
        u->sz = 0;
        for(long long i = 0; i < 2; ++ i) if(u->suc[i] != NULL) u->sz += u->suc[i]->sz;
        delete _u; return u;
    }
    inline void merge(node *u) {rt = merge(rt, u);}
    long long query(node *u)
    {
        if(! u->vst) return u->L;
        else if(u->suc[0] != NULL && u->suc[0]->sz) return query(u->suc[0]);
        else return query(u->suc[1]);
    }
    inline long long query() {return query(rt);}
}seg[2];
int main()
{

#ifndef ONLINE_JUDGE

    freopen("mex.in", "r", stdin);
    freopen("mex.out", "w", stdout);

#endif

    using namespace Zeonfai;
    seg[0].initialize();
    long long n = getInt();
    for(long long i = 0; i < n; ++ i)
    {
        long long opt = getInt(); long long L = getInt(), R = getInt();
        if(opt == 1) seg[1].merge(seg[0].split(L, R));
        else if(opt == 2) seg[0].merge(seg[1].split(L, R));
        else if(opt == 3)
        {
            segmentTree::node *u = seg[0].split(L, R), *v = seg[1].split(L, R);
            seg[0].merge(v); seg[1].merge(u);
        }
        printf("%lld\n", seg[0].query());
    }
}