CF EDU 103 C - Longest Simple Cycle

C - Longest Simple Cycle

dp

设 \(f[i]\) 为以第 \(i\) 个线段为环的右边界，环的最大长度

不妨令 \(l = min(a[i],b[i]),r =max(a[i],b[i])\)

1. \(l\neq r\)

   1. 第 \(i-1\) 条线段的 \([l,r]\) 这部分作为环的左边界，此时环的长度为 \(r-l+1+c[i]\)
   2. 不选第 \(i-1\) 条线段的 \([l,r]\) 这部分，与以第 $i-1 $ 条线段为右端点的环解上，此时环的长度为 \(f[i-1]-(r-l-1)+c[i]\)

2. \(l == r\)

   ​ 只有策略 1 可以选，\(f[i]=c[i]+1\)

#include <iostream>
#include <cstring>
#include <algorithm>
#include <vector>
#include <cmath>

using namespace std;
typedef long long ll;
const int N = 1e5 + 10;
int n;
ll c[N], a[N], b[N], f[N];
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
	int T;
	cin >> T;
	while(T--)
	{
		cin >> n;
		for (int i = 1; i <= n; i++)
			cin >> c[i];
		for (int i = 1; i <= n; i++)
			cin >> a[i];
		for (int i = 1; i <= n; i++)
			cin >> b[i];
		f[1] = 0;
		for (int i = 2; i <= n; i++)
		{
			ll l = min(a[i], b[i]), r = max(a[i], b[i]);
			if (l == r)
			{
				f[i] = c[i] + 1;
				continue;
			}
			f[i] = max(r - l + 1, f[i-1] - (r - l - 1)) + c[i];
		}
		cout << *max_element(f + 1, f + n + 1) << endl;
	}
	return 0;
}