Segment Tree Beats

\(Q1.\)給定長度為\(n\)的序列\(A\)，支援以下操作：1、區間取\(\min\)；2、區間查詢最大值；3、區間求和。

const int N = 1000005;
const int inf = 1<<30;
int n, m, a[N];
#define lc (o << 1)
#define rc (o << 1 | 1)
int ma[N*4], se[N*4], tot[N*4];
ll sum[N*4];
void pushup(int o) {
	sum[o] = sum[lc] + sum[rc];
	ma[o] = std::max(ma[lc], ma[rc]);
	se[o] = std::max(se[lc], se[rc]);
	if (ma[lc] != ma[rc]) chkmax(se[o], std::min(ma[lc], ma[rc]));
	tot[o] = (ma[o] == ma[lc] ? tot[lc] : 0) + (ma[o] == ma[rc] ? tot[rc] : 0);
}
void pushdown(int o) {
	if (ma[lc] > ma[o]) sum[lc] -= 1ll*(ma[lc]-ma[o])*tot[lc], ma[lc] = ma[o];
	if (ma[rc] > ma[o]) sum[rc] -= 1ll*(ma[rc]-ma[o])*tot[rc], ma[rc] = ma[o];
}
void build(int o, int l, int r) {
	if (l == r) { sum[o] = ma[o] = a[l], se[o] = -inf, tot[o] = 1; return; }
	int mid = l+r>>1;
	build(lc, l, mid), build(rc, mid+1, r);
	pushup(o);
}
void modify(int o, int l, int r, int x, int y, int v) {
	if (r < x || y < l) return;
	if (x <= l && r <= y && v > se[o]) {
		if (v >= ma[o]) return;
		sum[o] -= 1ll*(ma[o]-v)*tot[o]; ma[o] = v; return;
	}
	int mid = l+r>>1; pushdown(o);
	modify(lc, l, mid, x, y, v), modify(rc, mid+1, r, x, y, v);
	pushup(o);
}
int qmax(int o, int l, int r, int x, int y) {
	if (r < x || y < l) return -inf;
	if (x <= l && r <= y) return ma[o];
	int mid = l+r>>1; pushdown(o);
	return std::max(qmax(lc, l, mid, x, y), qmax(rc, mid+1, r, x, y));
}
ll qsum(int o, int l, int r, int x, int y) {
	if (r < x || y < l) return 0;
	if (x <= l && r <= y) return sum[o];
	int mid = l+r>>1; pushdown(o);
	return qsum(lc, l, mid, x, y) + qsum(rc, mid+1, r, x, y);
}
int main() {
	for (int T = read(); T; T--) {
		n = read(), m = read();
		rep(i, 1, n) a[i] = read();
		build(1, 1, n);
		while (m--) {
			int ty = read(), x = read(), y = read();
			if (!ty) { int t = read(); modify(1, 1, n, x, y, t); }
			else printf("%lld\n", ty == 1 ? qmax(1, 1, n, x, y) : qsum(1, 1, n, x, y));
		}
	}
	return 0;
}