Remove the Bracket

题目链接

题目描述：

RSJ has a sequence $a$ of $n$ integers $a_1,a_2,\dots ,a_n$ and $a_n$ integer $s$. For each of $a_2,a_3,\dots ,a_{n-1}$, he chose a pair of non-negative integers $x_i$ and $y_i$ such that $x_i+y_i=a_i$ and $(x_i-s)\cdot (y_i-s)\ge 0$.

Now he is interested in the value

$$F=a_1\cdot x_2+y_2\cdot x_3+y_3\cdot x_4+\dots +y_{n-2}\cdot x_{n-1}+y_{n-1}\cdot a_n.$$

Please help him find the minimum possible value $F$ he can get by choosing $x_i$ and $y_i$ optimally. It can be shown that there is always at least one valid way to choose them.

输入描述：

Each test contains multiple test cases. The first line contains an integer $t(1\le t\le {10}^4)$ — the number of test cases.

The first line of each test case contains two integers $n,s(3\le n\le 2\cdot {10}^5;0\le s\le 2\cdot {10}^5)$.

The second line contains $n$ integers $a_1,a_2,\dots ,a_n(0\le a_i\le 2\cdot {10}^5)$.

It is guaranteed that the sum of $n$ does not exceed $2\cdot {10}^5$.

输出描述：

For each test case, print the minimum possible value of $F$.

样例：

input:

10
5 0
2 0 1 3 4
5 1
5 3 4 3 5
7 2
7 6 5 4 3 2 1
5 1
1 2 3 4 5
5 2
1 2 3 4 5
4 0
0 1 1 1
5 5
4 3 5 6 4
4 1
0 2 1 0
3 99999
200000 200000 200000
6 8139
7976 129785 12984 78561 173685 15480

output:

0
18
32
11
14
0
16
0
40000000000
2700826806

Note:

In the first test case, $2\cdot 0+0\cdot 1+0\cdot 3+0\cdot 4=0$.

In the second test case, $5\cdot 1+2\cdot 2+2\cdot 2+1\cdot 5=18$.

AC代码：

题目中公式

$$F=a_1\cdot x_2+y_2\cdot x_3+y_3\cdot x_4+y_{n-2}\cdot x_{n-1}+y_{n-1}\cdot a_n$$

对于$\mathrm{\forall }$$i(2\le i\le n-1)$ 提取出

$$y_{i-1}\cdot x_i+y_i\cdot x_{i+1}$$

由于$x_i+y_i=a_i$, 所以$y_i=a_i-x_i$，带入原式中得

$$y_{i-1}\cdot x_i+(a_i-x_i)\cdot x_{i+1}$$

所以原式就变成了关于$x_i$的一次函数，那么函数的最值也就是会在$x_i$取最值的时候取到

• $a_i\le s$时，可以取最值$0$和$a_i$
• $a_i>s$时，可以取最值$s$和$a_i-s$

这两种情况都保证满足$(x_i-s)\cdot (y_i-s)\ge 0$

这样可以采取DP

采用一个数组$f[i][j]$，其状态表示为

• 当$j=0$时，$x_i$取最小值时的结果
• 当$j=1$时，$x_i$取最大值时的结果

而$f[i][j]$则有可能是从$f[i-1][0]$ 或 $f[j-1][1]$ 中转移过来，所以状态计算为

• $f[i][0]=min(f[i-1][0]+maxv(y_{i-1})\cdot minv(x_i),f[i-1][1]+minv\cdot minv)$
• $f[i][1]=min(f[i-1][0]+maxv(y_{i-1})\cdot maxv(x_i),f[i-1][1]+minv\cdot maxv)$

最后结果取$f[n][0]$和$f[n][1]$的最小值即可（最终结果$f[n][0]$和$f[n][1]$大小是一样的，但是我不明白为什么）

#include <bits/stdc++.h>
 
using namespace std;
 
typedef long long LL;
 
const int N = 2e5 + 10;
 
LL f[N][2];
LL x[N], y[N];
 
void solve()
{
	int n, s;
	cin >> n >> s;
 
	for(int i = 1; i <= n; i ++)
	{
		int m;
		cin >> m;
 
		if(i == 1 || i == n)
		{
			x[i] = y[i] = m;
		}
		else if(m <= s)
		{
			x[i] = 0, y[i] = m;
		}
		else
		{
			x[i] = min(s, m - s), y[i] = max(s, m - s);
		}
	}
 
	f[1][0] = f[1][1] = 0;
 
	for(int i = 2; i <= n; i ++)
	{
		f[i][0] = min(f[i - 1][0] + y[i - 1] * x[i], f[i - 1][1] + x[i - 1] * x[i]);
		f[i][1] = min(f[i - 1][0] + y[i - 1] * y[i], f[i - 1][1] + x[i - 1] * y[i]);
	}
	
	cout << f[n][0] << '\n';
}
 
int main()
{
	ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
 
	int T;
	cin >> T;
 
	while(T --)
		solve();
 
	return 0;
}