传送门

Description

You are given an array \([1,2,…,n]\), where the number of elements \(n\) is even.

In one operation, you can delete two adjacent elements of the array. If these elements are \(i\) and \(j\), the cost of this operation is \(cost(i, j)\).

In \(\frac{n}{2}\) operations, all elements will be deleted. The cost of deleting the whole array is defined as the largest cost among all the \(\frac{n}{2}\) operations.

What is the smallest possible cost of deleting the whole array?

It is guaranteed that the costs form a permutation of numbers from \(1\) to \({(\frac{n}{2})}^2\).

\(n\le 4000\)

Solution

设 \(dp(l，r)\) 表示删掉区间 \([l, r]\) 的答案。则转移为：

正常转移为\(O(n^3)\)。

考虑按照从小到大的顺序更新整个 \(dp\) 表。

考虑当前 \(dp(i,j) < val\) 的 \(dp\) 值都已经计算完毕，当前 \(cost(l, r) = val\)。

考虑哪些 \((i,j)\) 满足 \(dp(i,j) = val\)：

• 如果 \(dp(l,r)\) 目前未被计算，并且 \(dp(l + 1, r - 1)\) 已经被计算了或者 \(l + 1 = r\)，那么有 \(dp(l,r) = val\)。
• 如果 \(dp(l,r) = val\)，并且 \(dp(l - 1, r + 1)\) 尚未被计算同时 \(cost(l-1, r + 1) < dp(l ,r)\)，那么有 \(dp(l -1 , r + 1) = val\)。
• 如果\(dp(l,r) = val\)，并且 \(dp(l, r')\) 尚未被计算同时 \(dp(r + 1, r')\) 已经被计算过了（\(<val\)），那么有 \(dp(l, r') = dp(l,r)\)。对于这种转移，可以利用std::bitset来寻找转移的位置，获得常数优化。

每个区间被计算后，就用它来更新它可能更新到的区间，每个区间都只会被计算一次。

最后复杂度为 \(O(\frac{n^3}{64})\)。

为了加速运算，bitset可以只开 \(2000\)，因为对于每个点有 \(cost\) 的位置只有一半。

Code

#pragma GCC optimize(2)
#pragma GCC optimize(3)
#pragma GCC optimize("Ofast")
#pragma GCC optimize("inline")
#pragma GCC optimize("-fgcse")
#pragma GCC optimize("-fgcse-lm")
#pragma GCC optimize("-fipa-sra")
#pragma GCC optimize("-ftree-pre")
#pragma GCC optimize("-ftree-vrp")
#pragma GCC optimize("-fpeephole2")
#pragma GCC optimize("-ffast-math")
#pragma GCC optimize("-fsched-spec")
#pragma GCC optimize("unroll-loops")
#pragma GCC optimize("-falign-jumps")
#pragma GCC optimize("-falign-loops")
#pragma GCC optimize("-falign-labels")
#pragma GCC optimize("-fdevirtualize")
#pragma GCC optimize("-fcaller-saves")
#pragma GCC optimize("-fcrossjumping")
#pragma GCC optimize("-fthread-jumps")
#pragma GCC optimize("-funroll-loops")
#pragma GCC optimize("-fwhole-program")
#pragma GCC optimize("-freorder-blocks")
#pragma GCC optimize("-fschedule-insns")
#pragma GCC optimize("inline-functions")
#pragma GCC optimize("-ftree-tail-merge")
#pragma GCC optimize("-fschedule-insns2")
#pragma GCC optimize("-fstrict-aliasing")
#pragma GCC optimize("-fstrict-overflow")
#pragma GCC optimize("-falign-functions")
#pragma GCC optimize("-fcse-skip-blocks")
#pragma GCC optimize("-fcse-follow-jumps")
#pragma GCC optimize("-fsched-interblock")
#pragma GCC optimize("-fpartial-inlining")
#pragma GCC optimize("no-stack-protector")
#pragma GCC optimize("-freorder-functions")
#pragma GCC optimize("-findirect-inlining")
#pragma GCC optimize("-fhoist-adjacent-loads")
#pragma GCC optimize("-frerun-cse-after-loop")
#pragma GCC optimize("inline-small-functions")
#pragma GCC optimize("-finline-small-functions")
#pragma GCC optimize("-ftree-switch-conversion")
#pragma GCC optimize("-foptimize-sibling-calls")
#pragma GCC optimize("-fexpensive-optimizations")
#pragma GCC optimize("-funsafe-loop-optimizations")
#pragma GCC optimize("inline-functions-called-once")
#pragma GCC optimize("-fdelete-null-pointer-checks")
#include<bits/stdc++.h>
#define mkp make_pair
using namespace std;
#define MN 4005
int n, a[MN][MN];
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 << 3) + (x << 1) + ch - '0';
		ch = getchar();
	}
	return x*f;
}

#define se second
#define fi first

int f[MN][MN];
bitset<MN / 2> vis[MN], vis2[MN];
std::pair<int, int> pos[MN * MN / 4];

void upd(int l, int r) {
    bitset<MN / 2> tmp;
    if(r + 1 <= n) {
        tmp = (~vis[l]) & vis[r + 1];
        int k = tmp._Find_first();
        while(k != tmp.size()) {
            int kk = k * 2 + (l % 2 == 0);
            f[l][kk] = f[l][r];
            vis[l][kk / 2] = vis2[kk][l / 2] = 1;
            upd(l, kk);
            k = tmp._Find_next(k);
        }
    }
    if(l > 1) {
        tmp = (~vis2[r]) & vis2[l - 1];
        int k = tmp._Find_first();
        while(k != tmp.size()) {
            int kk = k * 2 + (r % 2 == 0);
            f[kk][r] = f[l][r];
            vis[kk][r / 2] = vis2[r][kk / 2] = 1;
            upd(kk, r);
            k = tmp._Find_next(k);
        }
    }
    if(l > 1 && r < n && !vis[l - 1][(r + 1) / 2] && a[l - 1][r + 1] < f[l][r]) {
        vis[l - 1][(r + 1) / 2] = 1;
        vis2[r + 1][(l - 1) / 2] = 1;
        f[l - 1][r + 1] = f[l][r];
        upd(l - 1, r + 1);
    }
}

int main() {
    n = read();
	for(int i = 1; i <= n; i++)
		for(int j = i + 1; j <= n; j += 2) {
            a[i][j] = read();
            pos[a[i][j]] = mkp(i, j);
        }
    int MAX = (n / 2) * (n / 2);
    for(int i = 1; i <= MAX; ++i) {
        int l = pos[i].fi, r = pos[i].se;
        if(vis[l][r / 2] == 0 && (vis[l + 1][(r - 1) / 2] == 1 || (r == l + 1))) {
            vis[l][r / 2] = vis2[r][l / 2] = 1;
            f[l][r] = i;
            upd(l, r);
        }
    }
    return 0 * printf("%d\n", f[1][n]);
}