显然地，所有数对可以构成一棵树。
按时间分治，设当前在处理 \([s,e]\) 内的修改，我们注意到每个修改会导致树上一条祖先后代链的数对被删除。
假设每次是从一个点 \(v\) 往上删到一点 \(u\)，我们不妨将所有 \(u\) 的父亲和所有 \(v\) 拿出来建出虚树，其余的边都可以扔掉。
途中需要用线段树来维护新增的数对。
直接实现就是 \(O((N+Q)\log^2(N+Q))\) 的。
然而事实上并不需要显式建树，注意到可以用单调栈来在按 DFS 序遍历的时候（也就是按左端点升序遍历）维护每个点的所有祖先。
可能会好写一点（大概）。
另，zkw 线段树在这题上很有用。
代码（从某份 std 那里抄来了 zkw）： 1
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using namespace std;
const int inf = 0x3f3f3f3f;
const int N = 1e5;
const int Q = 2.5e5;
int n, q;
int a[N + 5], las[N + 5];
pair<int, int> upd[N + Q + 5];
int pre[N + Q + 5], nxt[N + Q + 5];
struct SegmentTree {
static const int S = 1 << 17;
int seg[(S << 1) + 5];
void insert(int u, int k) {
seg[u += S] = k;
for (u >>= 1; u; u >>= 1)
seg[u] = max(seg[u << 1], seg[u << 1 | 1]);
}
int queryL(int u, int k) {
for (u += S; seg[u] < k && (u & (u - 1)); u = (u - 1) >> 1);
if (seg[u] < k) return 0;
for (; u < S; u |= seg[u | 1] >= k) u <<= 1;
return seg[u] == k ? u - S : 0;
}
int queryR(int u, int k) {
for (u += S; seg[u] < k && (u & (u + 1)); u = (u + 1) >> 1);
if (seg[u] < k) return n + 1;
for (; u < S; u |= seg[u] < k) u <<= 1;
return seg[u] == k ? u - S : n + 1;
}
} seg;
struct node {
int l, r, h, w;
node(int l = 0, int r = 0, int h = 0, int w = 0): l(l), r(r), h(h), w(w) {}
bool operator<(const node &o) const { return l < o.l; }
bool operator==(const node &o) const { return l == o.l; }
};
pair<vector<node>, int> build(vector<int> newStatic, vector<pair<int, int>> upd, vector<node> tr) {
sort(newStatic.begin(), newStatic.end()), sort(upd.begin(), upd.end());
static int anc[N + 5];
static node newTr[N * 2 + 5];
int j = 0, top = 0, tot = 0;
for (auto [i, _]: upd) seg.insert(i, 0);
for (int i: newStatic) {
seg.insert(i, a[i]);
for (; j < tr.size() && tr[j].l <= i; ++j) {
auto [l, r, h, w] = tr[j];
for (; top && tr[anc[top]].r <= r; --top) newTr[tot++] = tr[anc[top]];
anc[++top] = j;
}
for (; top && tr[anc[top]].r < i; --top) newTr[tot++] = tr[anc[top]];
for (; top && tr[anc[top]].h <= a[i]; --top);
}
for (; top; --top) newTr[tot++] = tr[anc[top]];
for (; j < tr.size(); ++j) newTr[tot++] = tr[j];
for (int i: newStatic) {
int l = seg.queryL(i - 1, a[i]);
if (l) newTr[tot++] = node(l, i, a[i], 1);
int r = seg.queryR(i + 1, a[i]);
if (r <= n) newTr[tot++] = node(i, r, a[i], 1);
}
tr = vector<node>(newTr, newTr + tot), sort(tr.begin(), tr.end()), tr.erase(unique(tr.begin(), tr.end()), tr.end());
vector<int> sum(tr.size());
int cnt = 0;
for (int i = 0; i < tr.size(); ++i) {
auto [l, r, h, w] = tr[i];
for (; top && tr[anc[top]].r <= r; --top);
if (top) sum[i] = sum[anc[top]];
sum[i] += w, cnt += w;
anc[++top] = i;
}
vector<int> crit(1, 0);
crit.reserve(upd.size() + 1);
j = top = 0;
for (auto [i, h]: upd) {
for (; j < tr.size() && tr[j].l <= i; ++j) {
auto [l, r, h, w] = tr[j];
for (; top && tr[anc[top]].r <= r; --top);
anc[++top] = j;
}
for (; top && tr[anc[top]].r < i; --top);
crit.push_back(anc[top]);
int l = 1, r = top, mid, res = 1;
while (l <= r) {
mid = (l + r) >> 1;
if (tr[anc[mid]].h > h) l = mid + 1, res = mid;
else r = mid - 1;
}
crit.push_back(anc[res]);
}
sort(crit.begin(), crit.end()), crit.erase(unique(crit.begin(), crit.end()), crit.end());
crit.reserve(crit.size() * 2);
j = top = 0;
for (int i = 0, siz = crit.size(); i + 1 < siz; ++i) {
for (; j <= crit[i]; ++j) {
auto [l, r, h, w] = tr[j];
for (; top && tr[anc[top]].r <= r; --top);
anc[++top] = j;
}
for (; top && tr[anc[top]].r <= tr[crit[i + 1]].r; --top);
crit.push_back(anc[top]);
}
sort(crit.begin(), crit.end()), crit.erase(unique(crit.begin(), crit.end()), crit.end());
top = 0;
for (int i = 0; i < crit.size(); ++i) {
auto [l, r, h, w] = tr[crit[i]];
for (; top && tr[anc[top]].r <= r; --top);
w = sum[crit[i]];
if (top) w -= sum[anc[top]];
cnt -= w;
anc[++top] = crit[i];
newTr[i] = node(l, r, h, w);
}
return {vector<node>(newTr, newTr + crit.size()), cnt};
}
int ans[N + Q + 5];
void solve(int l, int r, vector<node> tr) {
if (l == r) {
a[upd[l].first] = upd[l].second;
int tag = build({upd[l].first}, {}, tr).second;
ans[l] += tag, ans[l + 1] -= tag;
return ;
}
int mid = (l + r) >> 1;
vector<int> tmp;
vector<pair<int, int>> updTmp;
static bool vis[N + 5];
if (mid >= n) {
for (int i = mid + 1; i <= r; ++i) if (pre[i] < l) tmp.push_back(upd[i].first), vis[upd[i].first] = 1;
updTmp = vector<pair<int, int>>(upd + l, upd + mid + 1);
for (int i = l; i <= mid; ++i) if (pre[i] && pre[i] < l && !vis[upd[pre[i]].first]) updTmp.push_back(upd[pre[i]]);
for (int i: tmp) vis[i] = 0;
auto [L, lTag] = build(tmp, updTmp, tr);
ans[l] += lTag, ans[mid + 1] -= lTag;
vector<int>().swap(tmp), vector<pair<int, int>>().swap(updTmp);
solve(l, mid, L);
}
for (int i = l; i <= mid; ++i) if (nxt[i] > r) tmp.push_back(upd[i].first), vis[upd[i].first] = 1;
updTmp = vector<pair<int, int>>(upd + mid + 1, upd + r + 1);
for (int i = mid + 1; i <= r; ++i) if (pre[i] && pre[i] <= mid && !vis[upd[pre[i]].first]) updTmp.push_back(upd[pre[i]]);
for (int i: tmp) vis[i] = 0;
auto [R, rTag] = build(tmp, updTmp, tr);
ans[mid + 1] += rTag, ans[r + 1] -= rTag;
solve(mid + 1, r, R);
}
int main() {
scanf("%d%d", &n, &q);
a[0] = a[n + 1] = inf;
for (int i = 1; i <= n; ++i)
scanf("%d", a + i),
upd[i] = make_pair(i, a[i]), las[i] = i;
for (int i = n + 1; i <= n + q; ++i)
scanf("%d%d", &upd[i].first, &upd[i].second),
pre[i] = las[upd[i].first], las[upd[i].first] = i;
fill(las + 1, las + n + 1, n + q + 1);
for (int i = n + q; i; --i) nxt[i] = las[upd[i].first], las[upd[i].first] = i;
solve(1, n + q, {node(0, n + 1, inf, 0)});
for (int i = 1; i <= n + q; ++i) ans[i] += ans[i - 1];
for (int i = n; i <= n + q; ++i) printf("%d\n", ans[i]);
}