小y的序列

// Problem: 小y的序列
// Contest: NowCoder
// URL: https://ac.nowcoder.com/acm/contest/11186/C
// Memory Limit: 524288 MB
// Time Limit: 6000 ms
// 2022-02-25 11:33:22
// 
// Powered by CP Editor (https://cpeditor.org)

#include
using namespace std;

#define rep(i,l,r) for(int i=(l);i<=(r);i++)
#define per(i,l,r) for(int i=(l);i>=(r);i--)
#define ll long long
#define pii pair
#define mset(s,t) memset(s,t,sizeof(t))
#define mcpy(s,t) memcpy(s,t,sizeof(t))
#define fi first
#define se second
#define pb push_back
#define all(x) (x).begin(),(x).end()
#define SZ(x) ((int)(x).size())
#define mp make_pair

const ll mod = 1e9 + 7;

inline ll qmi (ll a, ll b) {
	ll ans = 1;
	while (b) {
		if (b & 1) ans = ans * a%mod;
		a = a * a %mod;
		b >>= 1;
	}
	return ans;
}
inline int read () {
	int x = 0, f = 0;
	char ch = getchar();
	while (!isdigit(ch)) f |= (ch=='-'),ch= getchar();
	while (isdigit(ch)) x = x * 10 + ch - '0', ch = getchar();
	return f?-x:x;
}
template void print(T x) {
	if (x < 0) putchar('-'), x = -x;
	if (x >= 10) print(x/10);
	putchar(x % 10 + '0');
}
inline ll sub (ll a, ll b) {
	return ((a - b ) %mod + mod) %mod;
}
inline ll add (ll a, ll b) {
	return (a + b) %mod;
}
inline ll inv (ll a) {
	return qmi(a, mod - 2);
}
const int N = 4e6 + 10;

struct Node {
	int l, r;
	int minv, maxv;
	int dif;
}t[N];
int n, k;
int a[N];

inline void push_up (int p) {
	t[p].maxv = max (t[p << 1].maxv, t[(p << 1)+ 1].maxv);
	t[p].minv =  min (t[p << 1].minv, t[(p << 1) + 1].minv);
}
inline void build (int p, int l, int r) {
	t[p].l = l, t[p].r = r;
	if (l == r) {
		t[p].l = l, t[p].r = r;
		t[p].maxv = t[p].minv = a[l];
		return;
	}
	int mid = l + r>> 1;
	build(p * 2, l, mid);
	build(p * 2 + 1, mid + 1, r);
	push_up(p);
}
inline void modify (int p, int x, int v) {
	if (t[p].l <= x && t[p].r >= x) {
		t[p].maxv = t[p].minv = v;
		return;
	}
	int mid = (t[p].l + t[p].r) >> 1;
	if (x <= mid) modify (p << 1, x, v);
	else modify ((p << 1) + 1, x, v );
	push_up (p);
}
inline pii query (int p, int l, int r) {
	if (t[p].l >= l && t[p].r <= r) {
		return mp(t[p].minv, t[p].maxv);
	}
	int val1=  0x3f3f3f3f, val2 = -0x3f3f3f3f;
	int mid = (t[p].l + t[p].r) >> 1;
	if (l <= mid) {
		  pii temp = query(p << 1, l, r);
		    val1 = min (val1, temp.fi);
			val2 = max(val2, temp.se);
	}
	if (r > mid)
	{		 pii temp = query((p <<1) + 1, l, r);
		    val1 = min (val1, temp.fi);
			val2 = max(val2, temp.se);
	}
	return mp(val1, val2);
}
// bool check1 (int i, int mid) {
	// int maxv = query(1, i, mid, 0);
	// int minv = query (1, i, mid, 1);
	// return maxv - minv >= k;
// }
// bool check2 (int i, int mid) {
	// int maxv = query(1, i, mid, 0);
	// int minv = query (1, i, mid, 1);
	// return maxv - minv <= k;
// }
// bool check3(int i, int mid) {
    // int maxv = query(1, i, mid, 0);
	// int minv = query (1, i, mid, 1);
	// return maxv - minv == k;
// }
void solve() {
    scanf("%d%d", & n, & k);
	rep(i, 1, n) 
        scanf("%d", & a[i]);
	    build(1, 1, n);
	     ll res = 0;
        int l = 1, r1 = 1, r2 = 1;
        for ( ; l <= n; l ++ ) {
                pair qry = query(1, l, r1);
                while ( r1 <= n && qry.second - qry.first < k ) r1 ++, qry = query(1, l, r1);
                qry = query(1, l, r2);
                while ( r2 <= n && qry.second - qry.first <= k ) r2 ++, qry = query(1, l, r2);
                r2 --;
                if ( r1 <= n ) res += (ll)r2 - (ll)r1 + 1;
                else break;
        }
        printf("%lldn", res);
}
int main () {
    int t;
    t =1;
    //cin >> t;
    while (t --) solve();
    return 0;
}

这一题要看到一个关键性质。固定一个左端，最大值和最小值的插值是递增的 用线段树O(Nlogn)

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