20240704

NFLSOJ

T2

CF1076G Array Game

阿瑞游戏。

考虑从后往前推。如果一个位置后面能走到的位置中有一个必败态，则一定会走过去，则当前位为必胜态。否则两人必然在当前位反复磨，直到一方不得不滚蛋为止。不得不滚蛋的那位就输了。因此当前位是否必胜只与当前位的奇偶性与后 \(m\) 位的状态有关。由于 \(m\) 很小，考虑直接压成一个数，对每一位记录后面 \(m\) 位的状态，转移相当于根据这些状态推出当前状态，然后把当前状态也扔到状态里，然后转移到（序列的）前一位。注意到这实际上是一个只与奇偶性有关的状态的映射，而且定义域很小。我们直接拿数组记下这个映射，放到线段树上维护。区间加法考虑若加的是偶数，则相当于没变。否则相当于把区间中所有数奇偶性取反。由于映射只有两种，我们对每个点都记两种映射，要取反就把两个映射交换。最后合并的时候注意映射合并的方向。

代码

#include <iostream>
#define int long long
using namespace std;
int n, m, q;
int a[200005];
struct f {short a[33];short& operator[](int x) { return a[x]; }
} T[800005][2];
f A[2];
f operator*(f a, f b) {f c;for (int i = 0; i < (1 << m); i++) c[i] = b[a[i]];return c;
}
struct Segment_Tree {bool tg[1000005];void tag(int o) {tg[o] ^= 1;swap(T[o][0], T[o][1]);}void pushdown(int o) {if (!tg[o]) return;tag(o << 1);tag(o << 1 | 1);tg[o] = 0;}void Build(int o, int l, int r) {if (l == r) {T[o][0] = A[a[l] & 1];T[o][1] = A[!(a[r] & 1)];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);T[o][0] = T[o << 1 | 1][0] * T[o << 1][0];T[o][1] = T[o << 1 | 1][1] * T[o << 1][1];}void Change(int o, int l, int r, int L, int R) {if (L <= l && r <= R) {tag(o);return;}pushdown(o);int mid = (l + r) >> 1;if (L <= mid) Change(o << 1, l, mid, L, R);if (R > mid) Change(o << 1 | 1, mid + 1, r, L, R);T[o][0] = T[o << 1 | 1][0] * T[o << 1][0];T[o][1] = T[o << 1 | 1][1] * T[o << 1][1];}f Query(int o, int l, int r, int L, int R) {if (L <= l && r <= R) return T[o][0];pushdown(o);int mid = (l + r) >> 1;if (R <= mid) return Query(o << 1, l, mid, L, R);if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);return Query(o << 1 | 1, mid + 1, r, L, R) * Query(o << 1, l, mid, L, R);}
} seg;
signed main() {cin >> n >> m >> q;for (int i = 1; i <= n; i++) cin >> a[i];for (short i = 0; i < (1 << m); i++) {if (i != (1 << m) - 1) A[0][i] = A[1][i] = (i << 1 | 1);else {A[0][i] = (i << 1 | 1);A[1][i] = (i << 1);}if (i & (1 << (m - 1)))  A[0][i] ^= (1 << m), A[1][i] ^= (1 << m);}seg.Build(1, 1, n);while (q--) {int op, l, r, d;cin >> op;if (op == 1) {cin >> l >> r >> d;if (d & 1)seg.Change(1, 1, n, l, r);} else {cin >> l >> r;f tmp = seg.Query(1, 1, n, l, r);cout << 2 - (tmp[(1 << m) - 1] & 1) << "\n";}}return 0;
}

T3

CF1681E Labyrinth Adventures

考虑在层之间做转移。注意到一个层的有效位置只有两个，也就是两个门的位置。层之间的转移都是线性的，可以直接使用矩阵刻画。询问的时候区间矩阵乘起来，再加上两头走到门的距离即可。

代码

#include <iostream>
#define int long long
using namespace std;
const int inf = 1000000000;
int n, m;
struct node {int x, y;
} d[100005][2];
struct Matrix {int a[2][2];int* operator[](int x) { return a[x]; }
} T[400005], I, A[400005];
Matrix operator*(Matrix a, Matrix b) {Matrix c;c[0][0] = min(a[0][0] + b[0][0], a[0][1] + b[1][0]);c[0][1] = min(a[0][0] + b[0][1], a[0][1] + b[1][1]);c[1][0] = min(a[1][0] + b[0][0], a[1][1] + b[1][0]);c[1][1] = min(a[1][0] + b[0][1], a[1][1] + b[1][1]);return c;
}
struct Segment_Tree {void Build(int o, int l, int r) {if (l == r) {T[o] = A[l];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);T[o] = T[o << 1] * T[o << 1 | 1];}Matrix Query(int o, int l, int r, int L, int R) {if (L <= l && r <= R) return T[o];int mid = (l + r) >> 1;if (R <= mid) return Query(o << 1, l, mid, L, R);if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);return Query(o << 1, l, mid, L, R) * Query(o << 1 | 1, mid + 1, r, L, R);}
} seg;
signed main() {cin >> n;for (int i = 1; i < n; i++) cin >> d[i][0].y >> d[i][0].x >> d[i][1].y >> d[i][1].x;for (int i = 1; i < n - 1; i++) {A[i][0][0] = (abs(d[i + 1][0].x - d[i][0].x));A[i][1][1] = (abs(d[i + 1][1].y - d[i][1].y));A[i][0][1] = (abs(d[i + 1][1].x - d[i][0].x)) + abs(d[i][0].y + 1 - d[i + 1][1].y);A[i][1][0] = (abs(d[i + 1][0].y - d[i][1].y)) + abs(d[i][1].x + 1 - d[i + 1][0].x);}if (n != 2) seg.Build(1, 1, n - 2);cin >> m;while (m--) {int x1, y1, x2, y2;cin >> y1 >> x1 >> y2 >> x2;int ca = max(x1, y1);int cb = max(x2, y2);if (ca > cb) swap(x1, x2), swap(y1, y2), swap(ca, cb);if (ca != cb) {--cb;int a = abs(x1 - d[ca][0].x) + abs(y1 - d[ca][0].y);int b = abs(x1 - d[ca][1].x) + abs(y1 - d[ca][1].y);int c = abs(x2 - d[cb][0].x) + abs(y2 - d[cb][0].y - 1) + 1;int e = abs(x2 - d[cb][1].x - 1) + abs(y2 - d[cb][1].y) + 1;I[0][0] = a, I[0][1] = b;I[1][0] = I[1][1] = inf;if (ca <= cb - 1) I = I * seg.Query(1, 1, n - 2, ca, cb - 1);cout << min(I[0][0] + c, I[0][1] + e) + cb - ca << "\n";} else cout << abs(x1 - x2) + abs(y1 - y2) << "\n";}return 0;
}

T4

CF1609E William the Obvilious

从前往后考虑，我们只关注当前匹配到 \(\texttt{abc}\) 的哪一位，因此把这个东西扔到状态里。转移都是线性的，矩乘刻画一下即可。

代码

#include <iostream>
#define int long long
using namespace std;
const int inf = 1000000000;
int n, q;
string str;
struct Matrix {int a[3][3];int* operator[](int x) { return a[x]; }
} M[3], T[400005], I;
Matrix operator*(Matrix a, Matrix b) {Matrix c;c[0][0] = min(a[0][0] + b[0][0], min(a[0][1] + b[1][0], a[0][2] + b[2][0]));c[0][1] = min(a[0][0] + b[0][1], min(a[0][1] + b[1][1], a[0][2] + b[2][1]));c[0][2] = min(a[0][0] + b[0][2], min(a[0][1] + b[1][2], a[0][2] + b[2][2]));c[1][0] = min(a[1][0] + b[0][0], min(a[1][1] + b[1][0], a[1][2] + b[2][0]));c[1][1] = min(a[1][0] + b[0][1], min(a[1][1] + b[1][1], a[1][2] + b[2][1]));c[1][2] = min(a[1][0] + b[0][2], min(a[1][1] + b[1][2], a[1][2] + b[2][2]));c[2][0] = min(a[2][0] + b[0][0], min(a[2][1] + b[1][0], a[2][2] + b[2][0]));c[2][1] = min(a[2][0] + b[0][1], min(a[2][1] + b[1][1], a[2][2] + b[2][1]));c[2][2] = min(a[2][0] + b[0][2], min(a[2][1] + b[1][2], a[2][2] + b[2][2]));return c;
}
struct Segment_Tree {void Build(int o, int l, int r) {if (l == r) {T[o] = M[str[l] - 'a'];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);T[o] = T[o << 1] * T[o << 1 | 1];}void Change(int o, int l, int r, int x) {if (l == r) {T[o] = M[str[x] - 'a'];return;}int mid = (l + r) >> 1;if (x <= mid) Change(o << 1, l, mid, x);else Change(o << 1 | 1, mid + 1, r, x);T[o] = T[o << 1] * T[o << 1 | 1];}
} seg;
signed main() {cin >> n >> q;cin >> str;str = ' ' + str;M[0][0][0] = 1,   M[0][0][1] = 0,   M[0][0][2] = inf;M[0][1][0] = inf, M[0][1][1] = 0,   M[0][1][2] = 1;M[0][2][0] = inf, M[0][2][1] = inf, M[0][2][2] = 0;M[1][0][0] = 0,   M[1][0][1] = 1,   M[1][0][2] = inf;M[1][1][0] = inf, M[1][1][1] = 1,   M[1][1][2] = 0;M[1][2][0] = inf, M[1][2][1] = inf, M[1][2][2] = 0;M[2][0][0] = 0,   M[2][0][1] = 1,   M[2][0][2] = inf;M[2][1][0] = inf, M[2][1][1] = 0,   M[2][1][2] = 1;M[2][2][0] = inf, M[2][2][1] = inf, M[2][2][2] = 1;seg.Build(1, 1, n);while (q--) {int x;char y;cin >> x >> y;str[x] = y;seg.Change(1, 1, n, x);I[0][0] = 0;I[0][1] = I[0][2] = inf;I[1][0] = I[1][1] = I[1][2] = inf;I[2][0] = I[2][1] = I[2][2] = inf;I = I * T[1];cout << min(I[0][0], min(I[0][1], I[0][2])) << "\n";}return 0;
}

T5

CF750E New Year and Old Subsequence

从前往后考虑，我们只关注当前匹配到 \(\texttt{2017}\) 和 \(\texttt{2016}\) 的哪一位。把这个东西扔到状态里，注意当匹配到 \(1\) 及其后的位时要对于 \(6\) 特殊转移，防止出现 \(\texttt{2016}\)。

代码

#include <iostream>
#define int long long
using namespace std;
const int inf = 2147483647;
int n, q;
string str;
struct Matrix {int a[5][5];int* operator[](int x) { return a[x]; }
} T[800005], A[10], I, E;
Matrix operator*(Matrix a, Matrix b) {Matrix c;for (int i = 0; i < 5; i++) {for (int j = 0; j < 5; j++) {c[i][j] = inf;for (int k = 0; k < 5; k++) c[i][j] = min(c[i][j], a[i][k] + b[k][j]);}}return c;
}
struct Segment_Tree {void Build(int o, int l, int r) {if (l == r) {T[o] = A[str[l] - '0'];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);T[o] = T[o << 1] * T[o << 1 | 1];}Matrix Query(int o, int l, int r, int L, int R) {if (L <= l && r <= R) return T[o];int mid = (l + r) >> 1;if (R <= mid) return Query(o << 1, l, mid, L, R);if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);return Query(o << 1, l, mid, L, R) * Query(o << 1 | 1, mid + 1, r, L, R);}
} seg;
signed main() {cin >> n >> q;cin >> str;str = ' ' + str;for (int i = 0; i < 5; i++) {for (int j = 0; j < 5; j++) E[i][j] = (i == j ? 0 : inf);}for (int i = 0; i < 10; i++) {A[i] = E;if (i != 0 && i != 1 && i != 2 && i != 6 && i != 7) continue;else if (i == 2) A[i][0][0] = 1, A[i][0][1] = 0;else if (i == 0) A[i][1][1] = 1, A[i][1][2] = 0;else if (i == 1) A[i][2][2] = 1, A[i][2][3] = 0;else if (i == 7) A[i][3][3] = 1, A[i][3][4] = 0;else A[i][3][3] = 1, A[i][4][4] = 1;}seg.Build(1, 1, n);while (q--) {int l, r;cin >> l >> r;I[0][0] = 0, I[0][1] = inf, I[0][2] = inf, I[0][3] = inf, I[0][4] = inf;I[1][0] = inf, I[1][1] = inf, I[1][2] = inf, I[1][3] = inf, I[1][4] = inf;I[2][0] = inf, I[2][1] = inf, I[2][2] = inf, I[2][3] = inf, I[2][4] = inf;I[3][0] = inf, I[3][1] = inf, I[3][2] = inf, I[3][3] = inf, I[3][4] = inf;I[4][0] = inf, I[4][1] = inf, I[4][2] = inf, I[4][3] = inf, I[4][4] = inf;I = I * seg.Query(1, 1, n, l, r);cout << (I[0][4] >= inf ? -1 : I[0][4]) << "\n";}return 0;
}

T6

CF1192B Dynamic Diameter

考虑动态 dp。先把边权下放到点权，设 \(f[i][0 / 1]\) 表示 \(i\) 子树内（包括 \(i\) 向上的边）的直径和还能向上延伸的最长链。重剖一下，将重儿子以外的 dp 值和当前点权视为常数，这些东西对重儿子 dp 值转移到当前点 dp 值的贡献可以使用矩阵刻画。这个矩阵可以根据转移方程列一下，应当是好列的。接下来考虑修改。首先对于一个点，如果知道其所有轻儿子的 dp 值，则可以根据当前点权构造出当前点的转移矩阵。也就是修改一个点实际上是修改了它的矩阵。这个矩阵更改后会直接影响到其所在链顶的 dp 值，所以这个链顶的 dp 值必须重新计算。由于链顶是轻儿子，而轻儿子的 dp 值变动会导致其父亲的转移矩阵变化，所以还需要重新计算链顶父亲的转移矩阵。然后依次类推，直到更新到根为止。重新计算某个点 dp 值的方法很简单，由于链底必为叶子，直接用这个叶子的初始矩阵按顺序乘上链上所有点的转移矩阵即可。线段树维护转移矩阵，需要注意矩阵乘法的方向。

代码

#include <iostream>
#include <set>
#define int long long
using namespace std;
const int inf = 4000000000000000000;
int n, q, W;
int head[100005], nxt[200005], to[200005], ew[200005], ecnt;
void add(int u, int v, int ww) { to[++ecnt] = v, nxt[ecnt] = head[u], head[u] = ecnt, ew[ecnt] = ww; }
int down[100005], _dfn[100005], son[100005], top[100005], dfn[100005], dep[100005], sz[100005], fa[100005], w[100005], ncnt;
int eu[100005], ev[100005];
int f[100005][2];
multiset<int, greater<int> > st[100005][2];
struct Matrix {int a[3][3];int* operator[](int x) { return a[x]; }Matrix() { a[0][0] = a[0][1] = a[0][2] = a[1][0] = a[1][1] = a[1][2] = a[2][0] = a[2][1] = a[2][2] = -inf; }
} A[100005], T[400005], E;
Matrix operator*(Matrix a, Matrix b) {Matrix c;c[0][0] = max(a[0][0] + b[0][0], max(a[0][1] + b[1][0], a[0][2] + b[2][0]));c[0][1] = max(a[0][0] + b[0][1], max(a[0][1] + b[1][1], a[0][2] + b[2][1]));c[0][2] = max(a[0][0] + b[0][2], max(a[0][1] + b[1][2], a[0][2] + b[2][2]));c[1][0] = max(a[1][0] + b[0][0], max(a[1][1] + b[1][0], a[1][2] + b[2][0]));c[1][1] = max(a[1][0] + b[0][1], max(a[1][1] + b[1][1], a[1][2] + b[2][1]));c[1][2] = max(a[1][0] + b[0][2], max(a[1][1] + b[1][2], a[1][2] + b[2][2]));c[2][0] = max(a[2][0] + b[0][0], max(a[2][1] + b[1][0], a[2][2] + b[2][0]));c[2][1] = max(a[2][0] + b[0][1], max(a[2][1] + b[1][1], a[2][2] + b[2][1]));c[2][2] = max(a[2][0] + b[0][2], max(a[2][1] + b[1][2], a[2][2] + b[2][2]));return c;
}
Matrix Construct(int x) {if (!son[x]) return E;int g0, g1 = *st[x][1].begin();if (sz[son[x]] + 1 == sz[x]) g0 = *st[x][0].begin();else {int tmp = *st[x][1].begin();st[x][1].erase(st[x][1].begin());g0 = max(*st[x][0].begin(), tmp + *st[x][1].begin());st[x][1].insert(tmp);}g1 += w[x];Matrix ret;ret[0][0] = ret[2][2] = 0;ret[1][0] = g1 - w[x];ret[2][1] = g1;ret[2][0] = g0;ret[1][1] = w[x];return ret;
}
void dfs1(int x, int fa, int d) {dep[x] = d;::fa[x] = fa;sz[x] = 1;for (int i = head[x]; i; i = nxt[i]) {int v = to[i];if (v != fa) {w[v] = ew[i];dfs1(v, x, d + 1);sz[x] += sz[v];if (sz[v] > sz[son[x]]) son[x] = v;}}
}
void dfs2(int x, int t) {top[x] = t;_dfn[dfn[x] = ++ncnt] = x;if (!son[x]) {down[t] = x;return;}dfs2(son[x], t);for (int i = head[x]; i; i = nxt[i]) {int v = to[i];if (v != fa[x] && v != son[x]) dfs2(v, v);}
}
void dfs3(int x) {f[x][1] = w[x];for (int i = head[x]; i; i = nxt[i]) {int v = to[i];if (v != fa[x]) {dfs3(v);f[x][0] = max(f[x][0], max(f[v][0], f[x][1] - w[x] + f[v][1]));f[x][1] = max(f[x][1], f[v][1] + w[x]);if (v != son[x]) {st[x][0].insert(f[v][0]);st[x][1].insert(f[v][1]);}}}if (son[x])A[dfn[x]] = Construct(x);else A[dfn[x]] = E;
}
struct Segment_Tree {void Build(int o, int l, int r) {if (l == r) {T[o] = A[l];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid);Build(o << 1 | 1, mid + 1, r);T[o] = T[o << 1 | 1] * T[o << 1];}void Change(int o, int l, int r, int x) {if (l == r) {T[o] = A[l];return;}int mid = (l + r) >> 1;if (x <= mid) Change(o << 1, l, mid, x);else Change(o << 1 | 1, mid + 1, r, x);T[o] = T[o << 1 | 1] * T[o << 1];}Matrix Query(int o, int l, int r, int L, int R) {if (L > R) return E;if (L <= l && r <= R) return T[o];int mid = (l + r) >> 1;if (R <= mid) return Query(o << 1, l, mid, L, R);if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);return  Query(o << 1 | 1, mid + 1, r, L, R) * Query(o << 1, l, mid, L, R);}
} seg;
void Change(int x) {if (son[x]) {A[dfn[x]] = Construct(x);seg.Change(1, 1, n, dfn[x]);}Matrix I;while (1) {int tmp = down[top[x]];I[0][0] = I[0][2] = 0, I[0][1] = w[tmp];I = I * seg.Query(1, 1, n, dfn[top[x]], dfn[tmp] - 1);x = top[x];int tmp0 = f[x][0], tmp1 = f[x][1];f[x][0] = I[0][0], f[x][1] = I[0][1];if (x == 1) break;st[fa[x]][0].erase(st[fa[x]][0].find(tmp0));st[fa[x]][1].erase(st[fa[x]][1].find(tmp1));st[fa[x]][0].insert(f[x][0]);st[fa[x]][1].insert(f[x][1]);x = fa[x];A[dfn[x]] = Construct(x);seg.Change(1, 1, n, dfn[x]);}
}
signed main() {E[0][0] = E[1][1] = E[2][2] = 0;cin >> n >> q >> W;for (int i = 1; i < n; i++) {int ww;int &u = eu[i], &v = ev[i];cin >> u >> v >> ww;add(u, v, ww);add(v, u, ww);}dfs1(1, 0, 1);dfs2(1, 1);dfs3(1);seg.Build(1, 1, n);int lans = 0;while (q--) {int d, ww;cin >> d >> ww;d = (d + lans) % (n - 1) + 1;ww = (ww + lans) % W;if (dep[eu[d]] < dep[ev[d]]) {w[ev[d]] = ww;Change(ev[d]);} else {w[eu[d]] = ww;Change(eu[d]);}cout << (lans = max(f[1][1], f[1][0])) << "\n";}return 0;
}

T10

Atcoder ABC228H Histogram

先排序，考虑最终的序列一定是一段一段，每一段都与最后一个数相同，而这最后一个数一直不变。因此直接 dp，发现可以斜率优化，然后就做完了。

代码

#include <iostream>
#include <algorithm>
#define int long long
using namespace std;
pair<int, int> a[200005];
int S[2][200005];
int n, X;
struct Line {int k, b;int operator()(int x) { return k * x + b; }
} stk[200005];
int sz;
bool chk(Line a, Line b, Line c) { return (b.b - a.b) * (b.k - c.k) >= (c.b - b.b) * (a.k - b.k); }
int dp[200005];
signed main() {cin >> n >> X;for (int i = 1; i <= n; i++) cin >> a[i].first >> a[i].second;sort(a + 1, a + n + 1);for (int i = 1; i <= n; i++) S[0][i] = S[0][i - 1] + a[i].second, S[1][i] = S[1][i - 1] + a[i].first * a[i].second;stk[++sz] = (Line) { 0, 0 };for (int i = 1, j = 1; i <= n; i++) {int x = a[i].first;while (j < sz && stk[j](x) >= stk[j + 1](x)) ++j;dp[i] = stk[j](x) - S[1][i] + a[i].first * S[0][i] + X;Line tmp = (Line) { -S[0][i], S[1][i] + dp[i] };while (sz > j && chk(stk[sz - 1], stk[sz], tmp)) --sz;stk[++sz] = tmp;}cout << dp[n] << "\n";return 0;
}
···
</details>### T11
CF1527E Partition Game直接线段树优化 dp 即可。也可以根据决策单调性分治。<details>
<summary> 代码 </summary>```cpp
#include <iostream>
#include <string.h>
#define int long long
using namespace std;
int n, k;
int f[35005][105];
int ap[35005];
int a[35005];
struct Segment_Tree {int mx[140005], tg[140005];void tag(int o, int v) { tg[o] += v, mx[o] += v; }void pushdown(int o) {if (!tg[o]) return;tag(o << 1, tg[o]);tag(o << 1 | 1, tg[o]);tg[o] = 0;}void Build(int o, int l, int r, int k) {tg[o] = 0;if (l == r) {mx[o] = f[l][k];return;}int mid = (l + r) >> 1;Build(o << 1, l, mid, k);Build(o << 1 | 1, mid + 1, r, k);mx[o] = min(mx[o << 1], mx[o << 1 | 1]);}void Add(int o, int l, int r, int L, int R, int v) {if (L <= l && r <= R) {tag(o, v);return;}pushdown(o);int mid = (l + r) >> 1;if (L <= mid) Add(o << 1, l, mid, L, R, v);if (R > mid) Add(o << 1 | 1, mid + 1, r, L, R, v);mx[o] = min(mx[o << 1], mx[o << 1 | 1]);}int Query(int o, int l, int r, int L, int R) {if (L <= l && r <= R) return mx[o];pushdown(o);int mid = (l + r) >> 1;if (R <= mid) return Query(o << 1, l, mid, L, R);if (L > mid) return Query(o << 1 | 1, mid + 1, r, L, R);return min(Query(o << 1, l, mid, L, R), Query(o << 1 | 1, mid + 1, r, L, R));}
} seg;
signed main() {cin >> n >> k;for (int i = 1; i <= n; i++) {cin >> a[i];if (ap[a[i]]) f[i][1] = f[i - 1][1] + i - ap[a[i]];else f[i][1] = f[i - 1][1];ap[a[i]] = i;}seg.Build(1, 1, n, 1);for (int x = 2; x <= k; x++) {memset(ap, 0, sizeof ap);f[1][x] = 0;ap[a[1]] = 1;for (int i = 2; i <= n; i++) {if (ap[a[i]] > 1) seg.Add(1, 1, n, 1, ap[a[i]] - 1, i - ap[a[i]]);ap[a[i]] = i;f[i][x] = seg.Query(1, 1, n, 1, i - 1);}seg.Build(1, 1, n, x);}cout << f[n][k] << "\n";return 0;
}