Educational Codeforces Round 102 (Rated for Div. 2) D（思维&线段树）

You are given a program that consists of nn instructions. Initially a single variable xx is assigned to 00. Afterwards, the instructions are of two types:

• increase xx by 11;
• decrease xx by 11.

 

You are given mm queries of the following format:

• query ll rr — how many distinct values is xx assigned to if all the instructions between the ll-th one and the rr-th one inclusive are ignored and the rest are executed without changing the order?

 

Input

The first line contains a single integer tt (1≤t≤10001≤t≤1000) — the number of testcases.

Then the description of tt testcases follows.

The first line of each testcase contains two integers nn and mm (1≤n,m≤2⋅1051≤n,m≤2⋅105) — the number of instructions in the program and the number of queries.

The second line of each testcase contains a program — a string of nn characters: each character is either '+' or '-' — increment and decrement instruction, respectively.

Each of the next mm lines contains two integers ll and rr (1≤l≤r≤n1≤l≤r≤n) — the description of the query.

The sum of nn over all testcases doesn't exceed 2⋅1052⋅105. The sum of mm over all testcases doesn't exceed 2⋅1052⋅105.

Output

For each testcase print mm integers — for each query ll, rr print the number of distinct values variable xx is assigned to if all the instructions between the ll-th one and the rr-th one inclusive are ignored and the rest are executed without changing the order.

Example

input

Copy

2
8 4
-+--+--+
1 8
2 8
2 5
1 1
4 10
+-++
1 1
1 2
2 2
1 3
2 3
3 3
1 4
2 4
3 4
4 4

output

Copy

1
2
4
4
3
3
4
2
3
2
1
2
2
2

Note

The instructions that remain for each query of the first testcase are:

1. empty program — xx was only equal to 00;
2. "-" — xx had values 00 and −1−1;
3. "---+" — xx had values 00, −1−1, −2−2, −3−3, −2−2 — there are 44 distinct values among them;
4. "+--+--+" — the distinct values are 11, 00, −1−1, −2−2.

 

题目大意：

给你一个数x，有n次操作，每个操作可以令x+1或者x-1，q次询问，每次询问你如果删除[l，r]的操作，x一共可以变成多少个数。

解法：

记max为过程中x可以成为的最大值，min为过程中x可以成为的最小值，答案就是max-min+1。

如果没有询问，变成的最大值或最小值就可以通过前缀和来实现，现在多了删除连续的区间，就要算它带来的贡献。

假设删除的区间为[L，R]，L之前的最大/小值和R之后的最大/小值都是可以通过线段树得到，不同的是，R之后的最大/小值要减去[L，R]的贡献即区间和才是删除后真正的最大/小值。

Accepted code

#pragma GCC optimize(3)
#include<bits/stdc++.h>
#include<unordered_map>
using namespace std;

#define sc scanf
#define Min(x, y) x = min(x, y)
#define Max(x, y) x = max(x, y)
#define ALL(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define pir pair <int, int>
#define MK(x, y) make_pair(x, y)
#define MEM(x, b) memset(x, b, sizeof(x))
#define MPY(x, b) memcpy(x, b, sizeof(x))
#define lowbit(x) ((x) & -(x))
#define P2(x) ((x) * (x))

typedef long long ll;
const int Mod = 1e9 + 7;
const int N = 2e5 + 100;
const int INF = 0x3f3f3f3f;
const ll LINF = 0x3f3f3f3f3f3f3f3f;
inline ll dpow(ll a, ll b){ ll r = 1, t = a; while (b){ if (b & 1)r = (r*t) % Mod; b >>= 1; t = (t*t) % Mod; }return r; }
inline ll fpow(ll a, ll b){ ll r = 1, t = a; while (b){ if (b & 1)r = (r*t); b >>= 1; t = (t*t); }return r; }

int mx[N * 4], mi[N * 4];
int pre[N], n, m;
char s[N];

#define ls (o << 1)
#define rs (ls | 1)
void Build(int o, int L, int R) {
	if (L == R) 
		mx[o] = mi[o] = pre[L];
	else {
		int mid = (L + R) >> 1;
		Build(ls, L, mid), Build(rs, mid + 1, R);
		mx[o] = max(mx[ls], mx[rs]);
		mi[o] = min(mi[ls], mi[rs]);
	}
}
int Ask_Mx(int o, int L, int R, int l, int r) {
	if (L >= l && R <= r)
		return mx[o];
	else {
		int mid = (L + R) >> 1, ans = -INF;
		if (mid >= l)
			Max(ans, Ask_Mx(ls, L, mid, l, r));
		if (mid < r)
			Max(ans, Ask_Mx(rs, mid + 1, R, l, r));
		return ans;
	}
}
int Ask_Mi(int o, int L, int R, int l, int r) {
	if (L >= l && R <= r)
		return mi[o];
	else {
		int mid = (L + R) >> 1, ans = INF;
		if (mid >= l)
			Min(ans, Ask_Mi(ls, L, mid, l, r));
		if (mid < r)
			Min(ans, Ask_Mi(rs, mid + 1, R, l, r));
		return ans;
	}
}

int main()
{
#ifdef OlaMins
	freopen("D:/input.txt", "r", stdin);
	//freopen("D:/output.txt", "w", stdout);
#endif

	int T; cin >> T;
	while (T--) {
		sc("%d %d %s", &n, &m, s + 1);
		for (int i = 1; i <= n; i++) 
			pre[i] = pre[i - 1] + (s[i] == '+' ? 1 : -1);  // 前缀和
		Build(1, 1, n);

		while (m--) {
			int l, r, mx1 = 0, mi1 = 0;
			sc("%d %d", &l, &r);

			if (l > 1) {
				Max(mx1, Ask_Mx(1, 1, n, 1, l - 1));   // L之前的答案可以直接得到
				Min(mi1, Ask_Mi(1, 1, n, 1, l - 1));
			}
			if (r < n) {    // R之后的答案要减去区间LR的贡献
				Max(mx1, Ask_Mx(1, 1, n, r + 1, n) - pre[r] + pre[l - 1]);
				Min(mi1, Ask_Mi(1, 1, n, r + 1, n) - pre[r] + pre[l - 1]);
			}

			printf("%d\n", mx1 - mi1 + 1);
		}
	}
	return 0; // 改数组大小!!!用pair改宏定义!!!
}