// Decline is inevitable,
// Romance will last forever.
#include <bits/stdc++.h>
using namespace std;
#define mst(a, x) memset(a, x, sizeof(a))
#define INF 0x3f3f3f3f
#define mp make_pair
#define pii pair<int,int>
//#define fi first
//#define se second
#define ll long long
#define LL long long
#define int long long
const int maxn = 5e5 + 10;
const int maxm = 1e3 + 10;
//const int P = 1e9 + 7;
int n, m;
struct segment_tree {
int l, r;
int sum, add;
int add_a, add_a1, add_b, add_b1;
int maxa, se, maxb, cnt;
#define l(x) tree[x].l
#define ls p<<1, l, mid
#define r(x) tree[x].r
#define rs p<<1|1, mid + 1, r
#define sum(x) tree[x].sum //区间和
#define add_a(x) tree[x].add_a //区间最大值的加法lazytag
#define add_a1(x) tree[x].add_a1 //区间非最大值的加法lazytag
#define add_b(x) tree[x].add_b //区间最大的历史最大值的加法lazytag
#define add_b1(x) tree[x].add_b1 //区间非最大的历史最大值的加法lazytag
#define maxa(x) tree[x].maxa //区间最大值
#define se(x) tree[x].se //区间严格次大值
#define maxb(x) tree[x].maxb //区间最大的历史最大值
#define cnt(x) tree[x].cnt //区间最大值个数
}tree[maxn << 2];
int a[maxn];
void push(int p) {
sum(p) = sum(p<<1) + sum(p<<1|1);
maxa(p) = max(maxa(p<<1), maxa(p<<1|1));
maxb(p) = max(maxb(p<<1), maxb(p<<1|1));
if(maxa(p<<1) == maxa(p<<1|1))
{
se(p) = max(se(p<<1), se(p<<1|1));
cnt(p) = cnt(p<<1) + cnt(p<<1|1);
}
else if(maxa(p<<1) > maxa(p<<1|1))
{
se(p) = max(se(p<<1), maxa(p<<1|1));
cnt(p) = cnt(p<<1);
}
else if(maxa(p<<1) < maxa(p<<1|1))
{
se(p) = max(maxa(p<<1), se(p<<1|1));
cnt(p) = cnt(p<<1|1);
}
}
void build(int p, int l, int r) {
l(p) = l; r(p) = r;
if(l == r) {
sum(p) = maxa(p) = maxb(p) = a[l];
se(p) = -INF;
cnt(p) = 1;
return;
}
int mid = (l + r) >> 1;
build(p<<1, l, mid);
build(p<<1|1, mid + 1, r);
push(p);
}
//所有对p的延迟标记都是对p的所有子节点生效,p点的数据标记前已经被修改了
void change(int p, int k1, int k2, int k3, int k4) {
//k1 最大值要加的数, k2 最大的历史最大值要加的数 k3 非最大值要加的数 k4非最大的历史最大值要加的数
sum(p) += 1LL * k1 * cnt(p) + 1LL * k3 * (r(p) - l(p) + 1 - cnt(p));
maxb(p)=max(maxb(p),maxa(p)+k2);
//如果当前区间的最大值加上最大的历史最大值要加的数大于原来的最大的历史最大值,就将值更新。
add_b(p)=max(add_b(p),add_a(p)+k2);
add_b1(p)=max(add_b1(p),add_a1(p)+k4);
maxa(p)+=k1;
add_a(p)+=k1;
add_a1(p)+=k3;
if(se(p)!=-INF) se(p)+=k3;
}
void spread(int p) {
int Max = max(maxa(p<<1), maxa(p<<1|1));
if(maxa(p<<1) == Max)
change(p<<1, add_a(p), add_b(p), add_a1(p), add_b1(p));
else
change(p<<1, add_a1(p), add_b1(p), add_a1(p), add_b1(p));
if(maxa(p<<1|1) == Max)
change(p<<1|1, add_a(p), add_b(p), add_a1(p), add_b1(p));
else
change(p<<1|1, add_a1(p), add_b1(p), add_a1(p), add_b1(p));
add_a(p) = add_b(p) = add_a1(p) = add_b1(p) = 0;
}
void add_change(int p, int l, int r, int d) {
if(l <= l(p) && r >= r(p)) {
change(p, d, d, d, d);
return ;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
if(l <= mid) add_change(p<<1, l, r, d);
if(r > mid) add_change(p<<1|1, l, r, d);
push(p);
}
void min_change(int p, int l, int r, int k) { //?
/*
if(l <= l(p) && r >= r(p) && k > se(p)) { //条件不要漏
change(p, k - maxa(p), k - maxa(p), 0, 0);
return;
}
spread(p);
int mid = (l(p) + r(p)) >> 1;
if(l <= mid) min_change(p<<1, l, r, k);
if(r > mid) min_change(p<<1|1, l, r, k);
push(p);
*/
if(l(p)>r||r(p)<l||k>=maxa(p))return;
if(l<=l(p)&&r(p)<=r&&k>se(p))
return change(p, k - maxa(p), k - maxa(p), 0, 0);
spread(p);
min_change(p<<1, l, r, k);
min_change(p<<1|1, l, r, k);
push(p);
}
int query(int p, int l, int r, int ope) { //index = 1, 2, 3对应区间求和,区间最大值,区间最大的历史最大值
if(ope == 1) {
if(l <= l(p) && r >= r(p)) return sum(p);
spread(p);
int mid = (l(p) + r(p)) >> 1;
int res = 0;
if(l <= mid) res += query(p<<1, l, r, ope);
if(r > mid) res += query(p<<1|1, l, r, ope);
return res;
}
else if(ope == 2) {
if(l <= l(p) && r >= r(p))
return maxa(p);
spread(p);
int mid = (l(p) + r(p)) >> 1;
int res = -2*INF;
if(l <= mid) res = max(res, query(p<<1, l, r, ope));
if(r > mid) res = max(res, query(p<<1|1, l, r, ope));
return res;
}
else {
if(l <= l(p) && r >= r(p)) return maxb(p);
spread(p);
int mid = (l(p) + r(p)) >> 1;
int res = -2*INF;
if(l <= mid) res = max(res, query(p<<1, l, r, ope));
if(r > mid) res = max(res, query(p<<1|1, l, r, ope));
return res;
}
}
void solve() {
cin >> n >> m;
for(int i = 1; i <= n; i++)
cin >> a[i];
build(1, 1, n);
while(m--) {
int ope, l, r;
cin >> ope >> l >> r;
if(ope == 1) {
int k;
cin >> k;
add_change(1, l, r, k);
}
else if(ope == 2) {
int k;
cin >> k;
min_change(1, l, r, k);
}
else {
cout << query(1, l, r, ope - 2) << endl;
}
}
}
signed main() {
ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
// int T; scanf("%d", &T); while(T--)
// freopen("1.txt","r",stdin);
// freopen("2.txt","w",stdout);
// int T; cin >> T;while(T--)
solve();
return 0;
}
/*
*/
P6242 【模板】线段树 3
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转载自blog.csdn.net/m0_59273843/article/details/120908329
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