转化一下，令 $a_x = y$，再做一遍这题，就可以过了。
这个转化挺显然的吧。

#include <bits/stdc++.h>
using namespace std;
const int N = 3e5 + 15;
int n, a[N], pos[N];
long long ans = 0;

struct Tree {
    int l, r;
    int Min, cnt;
    int flag;
} tr[N << 2];

void pushup(int u) {
    tr[u].Min = min(tr[u << 1].Min, tr[u << 1 | 1].Min);
    tr[u].cnt = 0;
    if (tr[u << 1].Min == tr[u].Min) tr[u].cnt += tr[u << 1].cnt;
    if (tr[u << 1 | 1].Min == tr[u].Min) tr[u].cnt += tr[u << 1 | 1].cnt;
}
void pushdown(int u) {
    if (tr[u].flag) {
        tr[u << 1].Min += tr[u].flag;
        tr[u << 1 | 1].Min += tr[u].flag;
        tr[u << 1].flag += tr[u].flag;
        tr[u << 1 | 1].flag += tr[u].flag;
        tr[u].flag = 0;
    }
}
void build(int u, int l, int r) {
    tr[u].l = l, tr[u].r = r;
    if (l == r) {
        tr[u].Min = 0, tr[u].cnt = 1;
        return;
    }
    int mid = l + r >> 1;
    build(u << 1, l, mid);
    build(u << 1 | 1, mid + 1, r);
    pushup(u);
}

void change(int u, int l, int r, int d) {
    if (tr[u].r < l || tr[u].l > r) return;
    if (tr[u].l >= l && tr[u].r <= r) {
        tr[u].Min += d, tr[u].flag += d;
        return;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1;
    if (l <= mid) change(u << 1, l, r, d);
    if (r > mid) change(u << 1 | 1, l, r, d);
    pushup(u);
}
int query(int u, int l, int r) {
    if (tr[u].r < l || tr[u].l > r) return 0;
    if (tr[u].l >= l && tr[u].r <= r) {
        if (tr[u].Min <= 1) return tr[u].cnt;
        return 0;
    }
    pushdown(u);
    int mid = tr[u].l + tr[u].r >> 1, res = 0;
    if (l <= mid) res += query(u << 1, l, r);
    if (r > mid) res += query(u << 1 | 1, l, r);
    return res;
}
bool st[N];

int main() {
    scanf("%d", &n);
    for (int i = 1, x, y; i <= n; i++) {
        scanf("%d%d", &x, &y);
        a[x] = y, pos[y] = x;
    }
    build(1, 1, n);
    for (int i = 1; i <= n; i++) {
        change(1, 1, i, 1);
        if (st[pos[i] - 1]) change(1, 1, a[pos[i] - 1], -1);
        if (st[pos[i] + 1]) change(1, 1, a[pos[i] + 1], -1);
        ans += query(1, 1, i);
        st[pos[i]] = 1;
    }
    printf("%lld\n", ans);
    return 0;
}