[IOI2014] Wall brick wall

Title Description

Given an initial element of \ (0 \) is the number of columns, and \ (K \) operations:

The interval \ ([L, R] \ ) in an element of \ (H \) take \ (max \)
The interval \ ([L, R] \ ) in an element of \ (H \) take \ (min \)

Problem-solving ideas

First, to be able to see that this is a problem in the tree line.
So how do we build a node of it?
First, for each node of the tree line, we remember the two markers \ (Min \) and \ (Max \) .
Because the topics related to \ (min \) and \ (max \) operation, so it should be difficult to think provided two such markers.
The significance of these two markers:

\ (Min [rt] \) represents the number \ (RT \) node comprises a section \ (min \) value of the flag
\ (Max [rt] \) represents the number \ (RT \) node comprises a section \ (max \) value of the flag

~~How it seems, and did not speak the same~~
Because the title last only asks each leaf node, so we do not care what the value of some node.
We just need to with each node and two marks: \ (Min \) and \ (Max \) , because we need both to update the leaves).
And these two markers can be updated by the interval to maintain.

How to deal with markers

Processing labeled with three operations: initialization, and downstream marking tag.
Simple analysis Initialization:
As our tag is used to update the child nodes (ie, the son of his father's mark \ (Max \) take \ (max \) , father of \ (Min \) take \ (min \) )
so we put \ (Max \) assigned to the minimum value, \ (Min \) assigned the maximum value:

Max[rt] = 0, Min[rt] = 0x3f3f3f3f;

Then look at marking:

inline void fMax(int rt, int h) { Min[rt] = max(Min[rt], h), Max[rt] = max(Max[rt], h); }

inline void fMin(int rt, int h) { Min[rt] = min(Min[rt], h), Max[rt] = min(Max[rt], h); }

~~In fact, these two functions are essentially the same~~
Let's analyze:
If we put a period to the interval \ (h \) take \ (max \) , then this section \ (Min \) mark and \ (Max \) tags should \ (h \) to take \ (max \) .
I do not make this specific analysis: Why do you think can be your own~~It can also be emotional understanding~~
So marking on the finished \ (QwQ \)

Finally, look at the pass mark \ (pushdown \)

inline void pushdown(int rt) {
    fMin(lc(rt), Min[rt]), fMin(rc(rt), Min[rt]);
    
    fMax(lc(rt), Max[rt]), fMax(rc(rt), Max[rt]);
    
    Max[rt] = 0, Min[rt] = 0x3f3f3f3f;  
}

In fact, this and similar marking, is to use his father's son updates.

How to answer output

This fact, as long traverse over the tree line, the leaf node \ (Max \) or \ (Min \) output can be.

Details Notes

Segment tree space open \ (4 \) times
Remember \ (pushdown \) after initialization also

Reference Code

/*--------------------------------
  Author: The Ace Bee
  Blog: www.cnblogs.com/zsbzsb
  This code is made by The Ace Bee
--------------------------------*/
#include <algorithm>
#include <iostream>
#include <cstring>
#include <cstdlib>
#include <cstdio>
#include <cctype>
#include <cmath>
#include <ctime>

#define rg register

using namespace std;

template < typename T > inline void read(T& s) {
    s = 0; int f = 0; char c = getchar();
    while (!isdigit(c)) f |= (c == '-'), c = getchar();
    while (isdigit(c)) s = s * 10 + (c ^ 48), c = getchar();
    s = f ? -s : s;
}

const int _ = 2000010;

int n, k, Min[_ << 2], Max[_ << 2];

inline int lc(int rt) { return rt << 1; }

inline int rc(int rt) { return rt << 1 | 1; }

inline void fMax(int rt, int h) { Min[rt] = max(Min[rt], h), Max[rt] = max(Max[rt], h); }

inline void fMin(int rt, int h) { Min[rt] = min(Min[rt], h), Max[rt] = min(Max[rt], h); }

inline void pushdown(int rt) {
    fMin(lc(rt), Min[rt]), fMin(rc(rt), Min[rt]);
    
    fMax(lc(rt), Max[rt]), fMax(rc(rt), Max[rt]);
    
    Max[rt] = 0, Min[rt] = 0x3f3f3f3f;
}

inline void update(int rt, int l, int r, int x, int y, int h, int t) {
    if (x <= l && r <= y) {
        if (t == 1)
            return fMax(rt, h);
        else
            return fMin(rt, h);
    }
    
    int mid = (l + r) >> 1;

    pushdown(rt);
    
    if (x <= mid) update(lc(rt), l, mid, x, y, h, t);
    
    if (y > mid) update(rc(rt), mid + 1, r, x, y, h, t);
}

inline void query(int rt, int l, int r) {
    if (l == r) { printf("%d\n", Max[rt]); return ; }
    
    int mid = (l + r) >> 1;
    
    pushdown(rt);
    
    query(lc(rt), l, mid);

    query(rc(rt), mid + 1, r);
}

int main() {
#ifndef ONLINE_JUDGE
    freopen("in.in", "r", stdin);
#endif
    read(n), read(k);

    for (rg int i = 1; i <= n << 2; ++i)
        Max[i] = 0, Min[i] = 0x3f3f3f3f;

    for (rg int t, l, r, h, i = 1; i <= k; ++i)
        read(t), read(l), read(r), read(h), update(1, 1, n, l + 1, r + 1, h, t);

    query(1, 1, n);
    
    return 0;
}

End Sahua \ (qwq \)