I. Reverse LIS The 7th China Collegiate Programming Contest, Finals (CCPC Finals 2021)

 写篇博客捋清一下思路

我们可以先算出一段区间的最优解将两个区间合并

对于一个区间记录4给状态vector[i][j] i，j属于{0,1}

代表以i开头，以j结尾的子序列状态，vector记录操作几次的最优状态

其中最优状态是经过诺干次操作后可以将字符串化成000(若干个0)11111(若干个1)

对于两个vector[i][j] vector[x][y]

合并成 vector[i][y]

一些细节就参考代码吧

参考166468081

//#include<bits/stdc++.h>
#include<iostream>
#include<algorithm>
#include<numeric>
#include<cstring>//rfind("string"),s.find(string,begin)!=s.npos,find_first _of(),find_last_of()
#include<string>//to_string(value),s.substr(int begin, int length);
#include<cstdio>
#include<cmath>
#include<vector>//res.erase(unique(res.begin(), res.end()), res.end()),resize(n)//size of vector,vector<int>().swap(at[mx])
#include<queue>//priority_queue(big)  /priority_queue<int, vector<int>, greater<int>> q(small)
#include<stack>
#include<map>
#include<set>
#include<unordered_map>
#include<unordered_set>
#include<bitset>
#include<random>
#include<chrono>
//#include<ext/pb_ds/assoc_container.hpp>//gp_hash_table
//#include<ext/pb_ds/hash_policy.hpp>
//using namespace __gnu_pbds;
std::mt19937_64 rnd(std::chrono::steady_clock::now().time_since_epoch().count());
using namespace std;
#define int long long//__int128 2^127-1(GCC)
#define PII pair<int,int>
const int inf = 0x3f3f3f3f3f3f3f3f, N = 1e5 + 5, mod = 1e9 + 7;
struct Info {
	vector<int>f[2][2];
	Info(){}
	Info(char x, int p) {
		if (x == '0') {
			f[0][0].push_back(p);
		}
		else {
			f[1][1].push_back(p);
		}
	}
};
void update(vector<int>& a,const vector<int>& b) {
	if (a.size() < b.size()) {
		a.resize(b.size(), -inf);
	}
	for (int i = 0; i < b.size(); i++) {
		a[i] = max(a[i], b[i]);
	}
}
vector<int> merge(const vector<int>& a, const vector<int>& b, int offset = 1) {
	if (a.empty() || b.empty()) {
		return{};
	}
	int i = 0, j = 0;
	vector<int>c{ a[0] + b[0] };
	if (offset) {
		c.insert(c.begin(), -inf);
	}
	while (i + 1 < a.size() && j + 1 < b.size()) {
		if (a[i + 1] - a[i] > b[j + 1] - b[j]) {
			c.push_back(c.back() + a[i + 1] - a[i]);
			i++;
		}
		else {
			c.push_back(c.back() + b[j + 1] - b[j]);
			j++;
		}
	}
	while (i + 1 < a.size()) {
		c.push_back(c.back() + a[i + 1] - a[i]);
		i++;
	}
	while (j + 1 < b.size()) {
		c.push_back(c.back() + b[j + 1] - b[j]);
		j++;
	}
	return c;
}
Info merge(const Info& a, const Info& b) {
	Info c = a;
	for (int i = 0; i < 2; i++) {
		for (int j = 0; j < 2; j++) {
			update(c.f[i][j], b.f[i][j]);
			for (int k = 0; k < 2; k++) {
				for (int l = 0; l < 2; l++) {
					update(c.f[i][l], merge(a.f[i][j], b.f[k][l], (j && !k)));
				}
			}
		}
	}
	return c;
}
signed main()
{
	ios_base::sync_with_stdio(0); cin.tie(0), cout.tie(0);
	int n;
	cin >> n;
	vector<char>c(n);
	vector<int>p(n);
	for (int i = 0; i < n; i++) {
		cin >> c[i] >> p[i];
	}
	auto solve = [&](auto solve, int l, int r)->Info {
		if (r - l == 1) {
			return Info(c[l], p[l]);
		}
		int m = l + r >> 1;
		auto res = merge(solve(solve, l, m), solve(solve, m, r));
		return res;
	};
	auto ans = solve(solve,0, n);
	int q;
	cin >> q;
	while (q--) {
		int k;
		cin >> k;
		int res = 0;
		for (int i = 0; i < 2; i++) {
			for (int j = 0; j < 2; j++) {
				int d = min(k, (int)ans.f[i][j].size() - 1);
				if (d >= 0) {
					res = max(res, ans.f[i][j][d]);
				}
			}
		}
		cout << res << "\n";
	}
}