BZOJ4364: [IOI2014]wall砖墙(线段树)

题意

题目链接

Sol

一个显然的思路是维护最大最小值以及最大最小值的覆盖标记。

https://paste.ubuntu.com/p/WXpBvzF6Y2/

但实际上因为这题只需要输出最后的操作序列，那么我们只维护最大最小值的覆盖标记即可。

也就是对于每一个节点，把本层的最大最小值下传之后清除即可。

// luogu-judger-enable-o2
#include<bits/stdc++.h> 
#define LL long long 
using namespace std;
const int MAXN = 8e6 + 10, INF = 1e9 + 10;
template <typename A, typename B> inline bool chmin(A &a, B b){if(a > b) {a = b; return 1;} return 0;}
template <typename A, typename B> inline bool chmax(A &a, B b){if(a < b) {a = b; return 1;} return 0;}
inline int read() {
    char c = getchar(); int x = 0, f = 1;
    while(c < '0' || c > '9') {if(c == '-') f = -1; c = getchar();}
    while(c >= '0' && c <= '9') x = x * 10 + c - '0', c = getchar();
    return x * f;
}
int N, K, opt[MAXN], L[MAXN], R[MAXN], H[MAXN];
#define ls k << 1
#define rs k << 1 | 1 
struct Node {
    int l, r, mx, mn;
}T[MAXN];
void psmin(int k, int v) {
    chmin(T[k].mx, v); chmin(T[k].mn, v);
}
void psmax(int k, int v) {
    chmax(T[k].mx, v); chmax(T[k].mn, v);
}
void pushdown(int k) {
    if(T[k].mn != INF)  psmin(ls, T[k].mn), psmin(rs, T[k].mn), T[k].mn = INF;
    if(T[k].mx != -INF) psmax(ls, T[k].mx), psmax(rs, T[k].mx), T[k].mx = -INF;
}
void Build(int k, int ll, int rr) {
    T[k].l = ll; T[k].r = rr; T[k].mn = INF; T[k].mx = -INF; 
    if(ll == rr) return ;
    int mid = ll + rr >> 1;
    Build(ls, ll, mid); 
    Build(rs, mid + 1, rr);
}
void Int(int k, int ll, int rr, int v, int opt) {
    if(ll <= T[k].l && T[k].r <= rr) {
        opt == 2 ? psmin(k, v) : psmax(k, v);
        return ;
    }
    pushdown(k);
    int mid = T[k].l + T[k].r >> 1;
    if(ll <= mid) Int(ls, ll, rr, v, opt);
    if(rr  > mid) Int(rs, ll, rr, v, opt);
}
void dfs(int k) {
    if(T[k].l == T[k].r) {printf("%d
", max(0, min(T[k].mn, T[k].mx)));return ;}
    pushdown(k);
    dfs(ls); dfs(rs);
}
signed main() {
    //freopen("a.in", "r", stdin);
    N = read(); K = read();
    Build(1, 1, N);
    for(int i = 1; i <= K; i++) {
        int opt = read(), L = read() + 1, R = read() + 1, H = read();
        if(opt == 1) Int(1, L, R, H, 1);
        else Int(1, L, R, H, 2);//Çø¼äÈ¡min 
    }
    dfs(1);
    return 0;
}
/*
6 3
1 0 4 9240
1 3 4 564
2 0 1 9249

6 3
1 1 5 9240
1 4 5 564
2 1 2 9249
*/