There is an apple tree in front of Taotao’s house. When autumn comes, n apples on the tree ripen, and Taotao will go to pick these apples.
When Taotao picks apples, Taotao scans these apples from the first one to the last one. If the current apple is the first apple, or it is strictly higher than the previously picked one, then Taotao will pick this apple; otherwise, he will not pick.
Given the heights of these apples h1,h2,⋯,hn, you are required to answer some independent queries. Each query is two integers p,q, which asks the number of apples Taotao would pick, if the height of the p-th apple were q (instead of hp). Can you answer all these queries?
Input
The first line of input is a single line of integer T (1≤T≤10), the number of test cases.
Each test case begins with a line of two integers n,m (1≤n,m≤105), denoting the number of apples and the number of queries. It is then followed by a single line of n integers h1,h2,⋯,hn (1≤hi≤109), denoting the heights of the apples. The next m lines give the queries. Each of these m lines contains two integers p (1≤p≤n) and q (1≤q≤109), as described in the problem statement.
Output
For each query, display the answer in a single line.
Sample Input
1
5 3
1 2 3 4 4
1 5
5 5
2 3
Sample Output
1
5
3
Hint
For the first query, the heights of the apples were 5, 2, 3, 4, 4, so Taotao would only pick the first apple.
For the second query, the heights of the apples were 1, 2, 3, 4, 5, so Taotao would pick all these five apples.
For the third query, the heights of the apples were 1, 3, 3, 4, 4, so Taotao would pick the first, the second and the fourth apples.
给定一个序列 ，现在有q 个操作，将 P位置上的数修改为x；
问每次修改后从首位置开始的最长递增子序列的长度；
思路：比较好的一道题；用单调栈维护一下； 说明一下各个变量：s1维护的是所选的序列的数字在原序列的下标，而 s2 是其值，stk 就是一个单调栈；另外注意的一点是，二分一定要会写，处理好边界，
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
#include<stack>
#include<functional>
#include<sstream>
using namespace std;
#define maxn 200005
#define inf 0x3f3f3f3f
#define INF 0x7fffffff
typedef long long ll;
typedef unsigned long long ull;
#define ms(x) memset(x,0,sizeof(x))
const long long int mod = 1e9 + 7;
inline int read()
{
int x = 0, k = 1; char c = getchar();
while (c < '0' || c > '9') { if (c == '-')k = -1; c = getchar(); }
while (c >= '0' && c <= '9')x = (x << 3) + (x << 1) + (c ^ 48), c = getchar();
return x * k;
}
int T, n, q;
int s1[maxn], s2[maxn], sq;
int ans[maxn];
int val[maxn];
int stk[maxn], bag;
int a[maxn];
struct node {
int pos, num, id;
}query[maxn];
bool cmp(node a, node b) {
return a.pos > b.pos;
}
int fin(int *gg, int l, int r, int x) {
if (gg[r] <= x)return r;
while (l < r) {
int mid = (l + r + 1) >> 1;
if (gg[mid] <= x)l = mid;
else r = mid - 1;
}
return l;
}
int main()
{
ios::sync_with_stdio(false);
int T; cin >> T;
while (T--) {
cin >> n >> q;
ms(ans); ms(s1); ms(s2); ms(val); ms(stk); ms(a);
bag = n; sq = 0;
int last = -1;
for (int i = 1; i <= n; i++) {
cin >> a[i];
if (last < a[i]) {
s1[++sq] = i; s2[sq] = a[i]; last = a[i];
}
val[i] = sq;
}
for (int i = 1; i <= q; i++) {
cin >> query[i].pos >> query[i].num; query[i].id = i;
}
s2[0] = int(-1e9); s1[0] = 0; s1[sq + 1] = s2[sq + 1] = int(1e9);
sort(query + 1, query + 1 + q, cmp);
for (int i = n, Q = 1; i >= 1;) {
if (query[Q].pos == i) {
int &res = ans[query[Q].id];
res = 0;
int left;
if (s1[sq] < query[Q].pos) {
left= sq;
}
else left = fin(s1, 0, sq, i);
if (s1[left] == i)left--;
res = left;
if (s2[left] < query[Q].num)res++;
else {
query[Q].num = s2[left];
}
if (stk[bag] > query[Q].num) {
res += n - bag + 1;
}
else {
if (stk[n] > query[Q].num) {
int pps = fin(stk, bag, n, query[Q].num);
res += n - pps;
}
}
Q++;
}
else {
if (stk[bag] > a[i]) {
stk[--bag] = a[i--]; continue;
}
int pos = fin(stk, bag, n, a[i]);
stk[pos] = a[i--]; bag = pos;// 更新栈
}
}
for (int i = 1; i <= q; i++) {
cout << ans[i] << endl;
}
}
return 0;
}