A. Flipping Game

A. Flipping Game

本质上是让我们找出一段区间内\(0\)的个数大于\(1\)的个数的最多的区间,且必须进行一次操作,所以可以考虑区间\(dp\),或者最小子序列和

1 最小子序列和

\[\begin{aligned} dp_i是以a_i结尾的最小子序列和 \\ dp_i=\min(dp_{i-1}+a[i],a[i]) \end{aligned} \]

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;
	vector<int> a(n + 1);
	i64 ans = 0 ;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		ans += a[i];
		if (!a[i]) a[i] = -1;
	}
	if (ans == n) ans --;
	vector<int> dp(n + 1);
	int k = 0;
	for (int i = 1; i <= n; i ++) {
		dp[i] = min(dp[i - 1] + a[i], a[i]);
		if (dp[i] < dp[k])
			k = i;
	}

	cout << ans - dp[k] << '\n';

	return 0;
}

2 区间dp

\[\begin{aligned} dp_{i,j}&为从i到j区间所有元素取反的和 \\ dp_{i,j}&=\max(dp_{i+1,j}+(1-a_i),dp_{i,j-1}+(1-a_j)) \end{aligned} \]

#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';

using namespace std;
using i64 = long long;

typedef pair<i64, i64> PII;

int main() {
	ios::sync_with_stdio(false);
	cin.tie(nullptr);

	int n;
	cin >> n;
	vector<int> a(n + 1), pre(n + 1);
	vector dp(n + 2, vector<int>(n + 2));
	int ans = 0 ;
	for (int i = 1; i <= n; i ++) {
		cin >> a[i];
		pre[i] = pre[i - 1] + a[i];
		dp[i][i] = a[i] ^ 1;
	}

	for (int len = 1; len <= n; len ++) {//枚举区间长度
		for (int i = 1; i + len - 1 <= n; i ++) {
			int j = i + len - 1;
			dp[i][j] = max(dp[i + 1][j] + (a[i] ^ 1), dp[i][j - 1] + (a[j] ^ 1));
			ans = max(ans, pre[i - 1] + dp[i][j] + pre[n] - pre[j]);
		}
	}

	cout << ans << '\n';

	return 0;
}