P8776 [蓝桥杯 2022 省 A] 最长不下降子序列
1.首先想到的做法
设up_len[i]为以a[i]为结尾的最长不下降子序列的长度，down_len[i]表示以a[i]为开始的最长不下降子序列的长度。
在求pre的过程中记录下额外信息：down_pre[i]表示在求down_len[i]的过程中，i是由哪个点转移过来的；
得到dp的转移方程：
if(down_pre[i]) ans = max(ans, up_len[i] + down_len[down_pre[i]] + min(k, down_pre[i] - i - 1)); else ans = max(ans, up_len[i] + min(k, n - i));
很好理解，这里采用了贪心
2.权值树状数组/权值线段树做法
参见这篇题解：
P8776 [蓝桥杯 2022 省 A] 最长不下降子序列 - 洛谷专栏 (luogu.com.cn)
！！！注意：
在最长不下降子序列中，
for (int i = 2; i <= n; i++) { if (a[i] >= en[res]) { en[++res] = a[i]; up_len[i] = res; } else { int pos = upper_bound(en + 1, en + res + 1, a[i]) - en; en[pos] = a[i]; up_len[i] = pos; } }
这里应该是upper_bound而不是lower_bound，因为最长不下降子序列的en数组可能会有连续若干个值是相同的，如果采用lower_bound的话，那么修改其中一个，其他相等的也要同时做出修改；而upper_bound因为找到的值是唯一的，只需要单独修改这个点的值即可
完整代码：
1.二分dp做法：
#include <iostream> #include <stdio.h> #include <algorithm> #include <string> #include <string.h> #include <cmath> #define R(x) x = read() #define For(i, j, n) for (int i = j; i <= n; ++i) using namespace std; const int N = 1e5 + 5; int n, k; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x * f; } int a[N], down_pre[N], up_len[N], down_len[N], en[N], id[N]; // id是反着做的辅助数组 int dp() { // 正着做 int res = 1, ans = -1; en[1] = a[1]; up_len[1] = 1; for (int i = 2; i <= n; i++) { if (a[i] >= en[res]) { en[++res] = a[i]; up_len[i] = res; } else { int pos = upper_bound(en + 1, en + res + 1, a[i]) - en; en[pos] = a[i]; up_len[i] = pos; } } // for(int i = 1; i <= n; i++) // cout << up_len[i] << " "; // 反着做 res = 1; memset(en, 0, sizeof en); down_len[n] = 1; en[res] = -a[n]; id[res] = n; for (int i = n - 1; i; i--) { int tmp = -a[i]; if (tmp >= en[res]) { down_pre[i] = id[res]; en[++res] = tmp; id[res] = i; down_len[i] = res; } else { int pos = upper_bound(en + 1, en + res + 1, tmp) - en; en[pos] = tmp; id[pos] = i; down_len[i] = pos; down_pre[i] = id[pos - 1]; } if(down_pre[i]) ans = max(ans, up_len[i] + down_len[down_pre[i]] + min(k, down_pre[i] - i - 1)); else ans = max(ans, up_len[i] + min(k, n - i)); } /*for (int i = 1; i <= n; i++) cout << down_len[i] << " " << down_pre[i] << endl;*/ ans = max(ans, up_len[n]); if(k >= n - 1) ans = n; return ans; } int main() { R(n); R(k); for (int i = 1; i <= n; i++) { R(a[i]); } printf("%d\n", dp()); return 0; }
2.权值树状数组（需要离散化）：
这种做法也用到了一个贪心：即修改k个数字，一定比修改少于k个数字更优
#include <iostream> #include <stdio.h> #include <algorithm> #include <string> #include <string.h> #include <cmath> #define RT register #define R(x) x = read() #define For(i, j, n) for (int i = j; i <= n; ++i) using namespace std; const int N = 1e5 + 5; int n, k; inline int read() { int x = 0, f = 1; char ch = getchar(); while (ch < '0' || ch > '9') { if (ch == '-') f = -1; ch = getchar(); } while (ch >= '0' && ch <= '9') { x = x * 10 + ch - '0'; ch = getchar(); } return x * f; } int a[N], down_pre[N], up_len[N], down_len[N], en[N], id[N]; // id是反着做的辅助数组 void dp() { // 正着做 int res = 1; en[1] = a[1]; up_len[1] = 1; for (int i = 2; i <= n; i++) { if (a[i] >= en[res]) { en[++res] = a[i]; up_len[i] = res; } else { int pos = lower_bound(en + 1, en + res + 1, a[i]) - en; en[pos] = a[i]; up_len[i] = pos; } } // for(int i = 1; i <= n; i++) // cout << up_len[i] << " "; // 反着做 res = 1; memset(en, 0, sizeof en); down_len[n] = 1; en[res] = -a[n]; id[res] = n; for (int i = n - 1; i; i--) { int tmp = -a[i]; if (tmp >= en[res]) { down_pre[i] = id[res]; en[++res] = tmp; id[res] = i; down_len[i] = res; } else { int pos = lower_bound(en + 1, en + res + 1, tmp) - en; en[pos] = tmp; down_len[i] = pos; down_pre[i] = id[pos - 1]; } } /*for (int i = 1; i <= n; i++) cout << down_len[i] << " " << down_pre[i] << endl;*/ } int tr[N], b[N], cnt; int lowbit(int x) { return x & (-x); } int query(int x) { int res = -1; while (x) { res = max(res, tr[x]); x -= lowbit(x); } return res; } void add(int x, int k) { for (int i = x; i <= n; i += lowbit(i)) tr[i] = max(tr[i], k); } int solve() { memcpy(b, a, sizeof(a)); sort(b + 1, b + n + 1); cnt = unique(b + 1, b + n + 1) - b - 1; for (int i = 1; i <= n; i++) a[i] = lower_bound(b + 1, b + n + 1, a[i]) - b; for (RT int i = 1; i <= n; i++) up_len[i] = query(a[i]) + 1, add(a[i], up_len[i]); memset(tr, 0, sizeof(tr)); for (RT int i = n; i >= 1; i--) down_len[i] = query(n - a[i] + 1) + 1, add(n - a[i] + 1, down_len[i]); memset(tr, 0, sizeof(tr)); int ans = 0; a[n + 1] = cnt + 1; for (int i = k + 2; i <= n + 1; i++) { add(a[i - k - 1], up_len[i - k - 1]); ans = max(ans, query(a[i]) + k + down_len[i]); // 注意这两行：a[i-k-1]和a[i],要理解权值树状数组的含义 } return ans; } int main() { R(n); R(k); if (k >= n - 1) { printf("%d\n", n); return 0; } for (int i = 1; i <= n; i++) { R(a[i]); } // dp(); printf("%d\n", solve()); return 0; }
