题意:
给你一个长度为n的数组,你现在可以进行两种操作:
1 p x:将第p个数的值修改为x。2 x:将数组中所有 <x的数全部修改为x。让你求q次操作后的数组。
思路:
首先这个题是单点修改+区间更新,那么一下就想到了线段树。对于区间更新我们肯定是会需要一个tag来标记,判断这个区间内的值是否会有变化的,所以线段树维护的值:
struct node
{
int l, r;
int minn, tag;
}tr[N << 2];
接下来就是单点更新和区间更新的过程了,还有就是向下维护的过程,需要更新minn和tag,然后这里我的区间和单点是分开修改的。
#include <bits/stdc++.h>
#define int long long
#define ul u << 1
#define ur u << 1 | 1
using namespace std;
const int N = 2e5 + 10;
struct node
{
int l, r;
int minn, tag;
}tr[N << 2];
int n, q;
void pushup(int u) { tr[u].minn = min(tr[ul].minn, tr[ur].minn); }
void pushdown(int u)
{
if (tr[u].tag != -1)
{
if (tr[ul].tag < tr[u].tag || tr[ul].tag == -1) tr[ul].tag = tr[u].tag;
if (tr[ur].tag < tr[u].tag || tr[ur].tag == -1) tr[ur].tag = tr[u].tag;
if (tr[ul].minn < tr[u].tag) tr[ul].minn = tr[u].tag;
if (tr[ur].minn < tr[u].tag) tr[ur].minn = tr[u].tag;
tr[u].tag = -1;
}
}
void build(int u, int l, int r)
{
tr[u].l = l, tr[u].r = r, tr[u].tag = -1;
if (l == r) { cin >> tr[u].minn; return ; }
int mid = l + r >> 1;
build(ul, l, mid), build(ur, mid + 1, r);
pushup(u);
}
void modify_point(int u, int p, int x)
{
if (tr[u].l == tr[u].r) { tr[u].minn = x; tr[u].tag = -1; return ; }
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
if (p <= mid) modify_point(ul, p, x);
else modify_point(ur, p, x);
pushup(u);
}
void modify_interval(int u, int l, int r, int x)
{
if (l <= tr[u].l && r >= tr[u].r)
{
if (tr[u].minn < x) { tr[u].minn = x; tr[u].tag = x; }
return ;
}
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
if (l <= mid) modify_interval(ul, l, r, x);
if (r > mid) modify_interval(ur, l, r, x);
pushup(u);
}
int query(int u, int x)
{
if (tr[u].l == tr[u].r) return tr[u].minn;
pushdown(u);
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) return query(ul, x);
else return query(ur, x);
}
signed main()
{
std::ios::sync_with_stdio(false);
cin >> n;
build(1, 1, n);
cin >> q;
int op, p, x;
while (q --)
{
cin >> op;
if (op == 1) { cin >> p >> x; modify_point(1, p, x); }
else { cin >> x; modify_interval(1, 1, n, x); }
}
for (int i = 1; i < n; i ++) cout << query(1, i) << ' ' ;
cout << query(1, n) << endl;
return 0;
}