P8862 「KDOI-03」还原数据

考虑操作逆序，那么对于操作 $1$，加法变成减法。

问题在于对于操作二，如何找到答案。

显然我们可以发现，经过这次操作二后，序列中的每个数都 $\geq$ 这次操作二的答案，于是操作二的答案必然 $\leq$ 这个操作二的区间最小值。

我们可以发现，答案应取区间最小值，因为假如答案可以小于区间最小值，那么取区间最小值时一定也可以。

线段树维护即可。

#include <bits/stdc++.h>
using namespace std;

#define int long long

const int N = 1e5 + 5, INF = 2e9, MOD = 1e9 + 7;

inline int read()
{
	int op = 1, x = 0;
	char ch = getchar();
	while ((ch < '0' || ch > '9') && ch != '-') ch = getchar();
	while (ch == '-')
	{
		op = -op;
		ch = getchar();
	}
	while (ch >= '0' and ch <= '9')
	{
		x = (x << 1) + (x << 3) + (ch ^ 48);
		ch = getchar();
	}
	return x * op;
}

inline void write(int x)
{
	if (x < 0) putchar('-'), x = -x;
	if (x > 9) write(x / 10);
	putchar(x % 10 + '0');
}

int n, q, t, a[N], b[N];

struct Modify
{
	int op, l, r, x;
	Modify(int _op, int _l, int _r, int _x): op(_op), l(_l), r(_r), x(_x){}
	Modify(){}
}G[N];

class SegmentTree
{
public:
	struct Node
	{
		int l, r, add, minn;
	}tr[N << 2];
	void pushup(int u)
	{
		tr[u].minn = min(tr[u << 1].minn, tr[u << 1 | 1].minn);
	}
	void pushdown(int u)
	{
		auto &rt = tr[u], &lf = tr[u << 1], &rit = tr[u << 1 | 1];
		if (rt.add != 0)
		{
			lf.add += rt.add;
			lf.minn += rt.add;
			rit.add += rt.add;
			rit.minn += rt.add;
			rt.add = 0;
		}
	}
	void build(int u, int l, int r)
	{
		tr[u] = { l, r, 0, b[l] };
		if (l == r) return;
		int mid = l + r >> 1;
		build(u << 1, l, mid);
		build(u << 1 | 1, mid + 1, r);
		pushup(u);
	}
	void update(int u, int l, int r, int x)
	{
		if (tr[u].l >= l and tr[u].r <= r)
		{
			tr[u].add += x;
			tr[u].minn += x;
			return;
		}
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1;
		if (l <= mid) update(u << 1, l, r, x);
		if (r > mid) update(u << 1 | 1, l, r, x);
		pushup(u);
	}
	int query(int u, int l, int r)
	{
		if (tr[u].l >= l and tr[u].r <= r)
		{
			return tr[u].minn;
		}
		pushdown(u);
		int mid = tr[u].l + tr[u].r >> 1, res = (int)1e16;
		if (l <= mid) res = query(u << 1, l, r);
		if (r > mid) res = min(res, query(u << 1 | 1, l, r));
		return res;
	}
}f;

signed main()
{
	// freopen("*.in", "r", stdin);
	// freopen("*.out", "w", stdout);
	scanf("%lld", &t);
	while (t--)
	{
		scanf("%lld%lld", &n, &q);
		for (int i = 1; i <= n; i++) scanf("%lld", &a[i]);
		for (int i = q; i >= 1; i--)
		{
			scanf("%lld%lld%lld", &G[i].op, &G[i].l, &G[i].r);
			if (G[i].op == 1) scanf("%lld", &G[i].x);
		}
		for (int i = 1; i <= n; i++) scanf("%lld", &b[i]);
		f.build(1, 1, n);
		vector<int> ans;
		for (int i = 1; i <= q; i++)
		{
			if (G[i].op == 1) f.update(1, G[i].l, G[i].r, -G[i].x);
			else 
			{
				ans.push_back(f.query(1, G[i].l, G[i].r));
			}
		}
		for (int i = ans.size() - 1; i >= 0; i--) printf("%lld ", ans[i]);
		printf("\n");
	}
	return 0;
}