Fun is Counting
我们可以发现数组 \(a\) 必须是 \(x\) 或 \(x - 1\),然后分类讨论即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 1e6 + 5, mod = 998244353;int inv[N], f[N], g[N], t, n, a[N];int C(int a, int b) {if (a < b) {return 0;}return f[a] % mod * g[b] % mod * g[a - b] % mod;
}void Solve() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i];}sort(a + 1, a + n + 1);if (a[1] + 1 < a[n]) {cout << "0\n";return ;}if (a[1] + 1 == a[n]) {int cnt = 0;for (int i = 1; i <= n; i++) {cnt += (a[i] == a[1]);}if (a[n] < cnt + 1) {cout << "0\n";return ;}cout << C(n, a[n]) * C(a[n], cnt) % mod * C(n - a[n] - 1, (a[n] - cnt) - 1) % mod << "\n";}else {if (a[1] == 1) {cout << n + (n == 2) << "\n";return ;}if (a[1] == n - 1) {cout << "1\n";return ;}cout << C(n - a[1] - 1, a[1] - 1) * C(n, a[1]) % mod << "\n";}
}signed main() {ios::sync_with_stdio(0);cin.tie(0);inv[1] = f[0] = g[0] = 1;for (int i = 1; i <= 1000000; i++) {f[i] = f[i - 1] * i % mod;if (i > 1) {inv[i] = inv[mod % i] * (mod - mod / i) % mod;}g[i] = g[i - 1] * inv[i] % mod;}cin >> t;while (t--) {Solve();}return 0;
}
Mad MAD Sum II
我们可以画出一个抽象的图,来表示他的值得变化
然后就可以想到一些更抽象的操作
那么我们就可以根据图所示,更改一段区间的值,有线段树维护即可
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 2e5 + 5, INF = 1e18;struct node {int x, sum, lazy;
}tr[N * 4];int t, n, a[N];map<int, int> mp;node Merge(node x, node y) {return {min(x.x, y.x), x.sum + y.sum, INF};
}void pushdown(int i, int l, int r, int mid) {if (tr[i].lazy != INF) {tr[i * 2] = {tr[i].lazy, (tr[i].lazy) * (mid - l + 1), tr[i].lazy};tr[i * 2 + 1] = {tr[i].lazy, (tr[i].lazy) * (r - mid), tr[i].lazy};}tr[i].lazy = INF;
}void modify(int i, int l, int r, int x, int y, int z) {if (l > y || r < x) {return ;}if (l >= x && r <= y) {tr[i] = {z, z * (r - l + 1), z};return ;}int mid = (l + r) >> 1;pushdown(i, l, r, mid);modify(i * 2, l, mid, x, y, z);modify(i * 2 + 1, mid + 1, r, x, y, z);tr[i] = Merge(tr[i * 2], tr[i * 2 + 1]);
}void build(int i, int l, int r) {if (l == r) {tr[i] = {INF, 0, INF};return ;}int mid = (l + r) >> 1;build(i * 2, l, mid);build(i * 2 + 1, mid + 1, r);tr[i] = {INF, 0, INF};
}int query(int i, int l, int r, int x, int y) {if (l > y || r < x) {return 0;}if (l >= x && r <= y) {return tr[i].sum;}int mid = (l + r) >> 1;pushdown(i, l, r, mid);return query(i * 2, l, mid, x, y) + query(i * 2 + 1, mid + 1, r, x, y);
}int Find(int i, int l, int r, int x) {if (l == r) {return l;}int mid = (l + r) >> 1;pushdown(i, l, r, mid);if (tr[i * 2].x > x) {return Find(i * 2 + 1, mid + 1, r, x);}return Find(i * 2, l, mid, x);
}void Solve() {cin >> n;for (int i = 1; i <= n; i++) {cin >> a[i];}build(1, 1, n);mp.clear();int ans = 0;for (int i = 1; i <= n; i++) {modify(1, 1, n, i, i, 0);int k = Find(1, 1, n, a[i]);if (k <= mp[a[i]]) {modify(1, 1, n, k, mp[a[i]], a[i]);}mp[a[i]] = i;ans += query(1, 1, n, 1, i);}cout << ans << "\n";
}signed main() {cin >> t;while (t--) {Solve();}return 0;
}
Not An SQRT Problem
说实话,就是更改一下 \(dfs\) 序即可,我们可以先更新儿子节点的 \(dfs\) 序,然后再递归
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 3e5 + 5, INF = 1e18;struct node {int x, lazy;
}tr[N * 4];int n, q, dfn[N], cnt[N], sz[N], dcnt, first[N];vector<int> g[N];void dfs1(int u, int f) {sz[u] = 1;for (auto v : g[u]) {if (v == f) {continue;}dfs1(v, u);sz[u] += sz[v];}
}void dfs2(int u, int f) {for (auto v : g[u]) {if (v == f) {continue;}if (!first[u]) {first[u] = v;}dfn[v] = ++dcnt;cnt[u]++;}for (auto v : g[u]) {if (v == f) {continue;}dfs2(v, u);}
}void pushdown(int i) {tr[i * 2].x += tr[i].lazy;tr[i * 2].lazy += tr[i].lazy;tr[i * 2 + 1].x += tr[i].lazy;tr[i * 2 + 1].lazy += tr[i].lazy;tr[i].lazy = 0;
}node Merge(node x, node y) {return {max(x.x, y.x), 0};
}void modify(int i, int l, int r, int x, int y, int z) {if (l > y || r < x) {return ;}if (l >= x && r <= y) {tr[i].x += z;tr[i].lazy += z;return ;}int mid = (l + r) >> 1;pushdown(i);modify(i * 2, l, mid, x, y, z);modify(i * 2 + 1, mid + 1, r, x, y, z);tr[i] = Merge(tr[i * 2], tr[i * 2 + 1]);
}int query(int i, int l, int r, int x, int y) {if (l > y || r < x) {return -INF;}if (l >= x && r <= y) {return tr[i].x;}int mid = (l + r) >> 1;pushdown(i);return max(query(i * 2, l, mid, x, y), query(i * 2 + 1, mid + 1, r, x, y));
}signed main() {ios::sync_with_stdio(0);cin.tie(0);cin >> n >> q;for (int i = 1, u, v; i < n; i++) {cin >> u >> v;g[u].push_back(v);g[v].push_back(u);}dfs1(1, 0);dfn[1] = 1;dcnt = 1;dfs2(1, 0);while (q--) {int opt;cin >> opt;if (opt == 2) {int x, v;cin >> x >> v;int cur = first[x];modify(1, 1, n, dfn[cur], dfn[cur] + cnt[x] - 1, v);}else if (opt == 1) {int x, v;cin >> x >> v;modify(1, 1, n, dfn[x], dfn[x], v);if (!g[x].empty()) {int cur = first[x];modify(1, 1, n, dfn[cur], dfn[cur] + sz[x] - 2, v);}}else if (opt == 3) {int x;cin >> x;int ans = query(1, 1, n, dfn[x], dfn[x]);if (!g[x].empty()) {int cur = first[x];ans = max(ans, query(1, 1, n, dfn[cur], dfn[cur] + sz[x] - 2));}cout << ans << '\n';}else {int x;cin >> x;int cur = first[x];cout << query(1, 1, n, dfn[cur], dfn[cur] + cnt[x] - 1) << "\n";}}return 0;
}
01 on Tree
和一战到底或者魔塔一模一样
#include <bits/stdc++.h>using namespace std;#define int long longconst int N = 2e5 + 5;struct node {int u, a, b;bool operator < (const node &_x) const {return b * _x.a > _x.b * a;}
};int n, ans, fa[N], f[N], a[N], b[N];priority_queue<node> q;int find(int x) {if (fa[x] == x) {return x;}return fa[x] = find(fa[x]);
}signed main() {cin >> n;for (int i = 2, u; i <= n; i++) {cin >> u;f[i] = u;}for (int i = 1, s; i <= n; i++) {cin >> s;a[i] = (s == 0);b[i] = (s == 1);q.push({i, (s == 0), (s == 1)});fa[i] = i;}while (!q.empty()) {if (q.top().u == 1 || a[q.top().u] != q.top().a || b[q.top().u] != q.top().b) {q.pop();continue;}int u = q.top().u, a1 = q.top().a, b1 = q.top().b;q.pop();int father = find(f[u]);ans += b[father] * a1;a[father] += a1;b[father] += b1;fa[u] = f[u];q.push({father, a[father], b[father]});}cout << ans;return 0;
}
