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:
- empty program — xx was only equal to 00;
- "-" — xx had values 00 and −1−1;
- "---+" — xx had values 00, −1−1, −2−2, −3−3, −2−2 — there are 44 distinct values among them;
- "+--+--+" — 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!!!
}