CodeForces - 833B The Bakery
Description
给一个长度为 \(n(n\le35000)\) 的序列,值域是 \([1,n]\) ,将它分成 \(k(k\le min(n,50))\) 段,求最大得分。定义每段的得分为这段的不同数个数。
Solution
令 \(dp[i][j]\) 表示前 \(i\) 个数分成 \(j\) 段的最大得分, \(num[i][j]\) 为 \([i,j]\) 中不同数个数。显然有 \(dp[i][j] = max\{dp[t][j-1]+num[t+1][i]\}\) 。每当我们加入一个数 \(a[i]\) ,则只有 \(num[pre[i]+1][i]\) 到 \(num[i][i]\) 会被加 \(1\) 。于是用线段树维护 \(dp[j][t]+cnt[j + 1][t]\) 即可。每次重新建树。
#include<bits/stdc++.h>
using namespace std;
template <class T> inline void read(T &x) {
x = 0; static char ch = getchar(); for (; ch < '0' || ch > '9'; ch = getchar());
for (; ch >= '0' && ch <= '9'; ch = getchar()) (x *= 10) += ch - '0';
}
#define N 35001
#define rep(i, a, b) for (int i = a; i <= b; i++)
#define ll long long
int n, K, a[N], pos[N], pre[N], dp[N], Max[N << 2], tag[N << 2];
#define ls rt << 1
#define rs ls | 1
#define mid (l + r >> 1)
#define pushUp Max[rt] = max(Max[ls], Max[rs])
void build(int rt, int l, int r) {
tag[rt] = 0;
if (l == r) { Max[rt] = dp[l]; return; }
build(ls, l, mid), build(rs, mid + 1, r);
pushUp;
}
inline void pushDown(int rt) {
int& t = tag[rt];
if (t) Max[ls] += t, Max[rs] += t, tag[ls] += t, tag[rs] += t, t = 0;
}
void update(int rt, int l, int r, int L, int R) {
if (L <= l && r <= R) { Max[rt]++, tag[rt]++; return; }
pushDown(rt);
if (L <= mid) update(ls, l, mid, L, R);
if (R > mid) update(rs, mid + 1, r, L, R);
pushUp;
}
int query(int rt, int l, int r, int L, int R) {
if (L <= l && r <= R) return Max[rt];
pushDown(rt); int ans = 0;
if (L <= mid) ans = max(ans, query(ls, l, mid, L, R));
if (R > mid) ans = max(ans, query(rs, mid + 1, r, L, R));
return ans;
}
int main() {
read(n), read(K);
rep(i, 1, n) read(a[i]), pre[i] = pos[a[i]], pos[a[i]] = i;
rep(j, 1, K) {
build(1, 0, n);
rep(i, 1, n) update(1, 0, n, pre[i], i - 1), dp[i] = query(1, 0, n, 0, i - 1);
}
printf("%d", dp[n]);
return 0;
}