题目链接
求给出的区间中有多少个三元组满足i+1=j=k-1 && a[i]<=a[j]<=a[k] 如果两个三元组的a[i], a[j], a[k]都相等， 那么这两个三元组算一个。
预处理一下所有三元组， 然后跑莫队就水过去了...
#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <complex>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef complex <double> cmx;
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
const int maxn = 2e5+5;
struct node
{
int l, r, id, block;
bool operator < (node a) const
{
if(block == a.block)
return r<a.r;
return block<a.block;
}
}q[maxn];
int a[maxn], b[maxn];
int n, m, ans[maxn], num[maxn];
map <pair<pll, int>, int> mp;
void solve() {
cin>>n;
mem(num);
mem(b);
int block = sqrt(n);
for(int i = 1; i <= n; i++) {
scanf("%d", &a[i]);
}
cin>>m;
for(int i = 0; i < m; i++) {
scanf("%d%d", &q[i].l, &q[i].r);
q[i].l++, q[i].r--;
q[i].block = q[i].l/block;
q[i].id = i;
}
sort(q, q+m);
int cnt = 0;
mp.clear();
for(int i = 2; i <= n-1; i++) {
if(a[i]>=a[i-1] && a[i]<=a[i+1]) {
if(!mp[mk(mk(a[i-1], a[i]), a[i+1])]) {
mp[mk(mk(a[i-1], a[i]), a[i+1])] = ++cnt;
}
b[i] = mp[mk(mk(a[i-1], a[i]), a[i+1])];
}
}
int L = 1, R = L-1, tot = 0;
for(int i = 0; i < m; i++) {
if(q[i].l>q[i].r) {
ans[q[i].id] = 0;
continue;
}
while(L<q[i].l) {
num[b[L]]--;
if(num[b[L]]==0&&b[L]!=0)
tot--;
L++;
}
while(L>q[i].l) {
L--;
num[b[L]]++;
if(num[b[L]]==1&&b[L]!=0)
tot++;
}
while(R>q[i].r) {
num[b[R]]--;
if(num[b[R]]==0&&b[R]!=0)
tot--;
R--;
}
while(R<q[i].r) {
R++;
num[b[R]]++;
if(num[b[R]]==1&&b[R]!=0)
tot++;
}
ans[q[i].id] = tot;
}
for(int i = 0; i < m; i++) {
printf("%d
", ans[i]);
}
}
int main()
{
int t;
cin>>t;
while(t--) {
solve();
}
return 0;
}