凸四边形不等式法则（平行四边形法则）

参考资料：四边形不等式优化讲解（详解）
前提条件:

1. \(dp_{i,j}=\max/\min\{dp_{i,k}+dp_{k+1,j}\}+cost_{i,j}\)
2. \(cost_{i,j}-cost_{i-1,j}\)单调递增

我们来证明几个性质：

1. 令\(i<i+1 \le j < j+1\)，则\(cost_{i,j}+cost_{i+1,j+1} \le cost_{i+1,j}+cost_{i,j+1}\)（即满足四边形不等式法则）
   证明：对原式进行移项，得\(cost_{i,j}-cost_{i+1,j} \le cost_{i,j+1}-cost_{i+1,j+1}\)

2. \(\forall i<i+1 \le j < j+1\)，则\(dp_{i,j}+dp_{i+1,j+1} \le dp_{i+1,j}+dp_{i,j+1}\)（既满足四边形不等式法则）
   证明：令\(dp_{i+1,j}\)在\(k=x\)时取最小值，\(dp_{i,j+1}\)在\(k=y\)时取最小值。(\(x<y\))
   \(dp_{i,j}+dp_{i+1,j+1}=dp_{i,x}+dp_{x+1,j}+cost_{i,j}+dp_{i+1,y}+dp_{y+1,j+1}+cost_{i+1,j+1}\)
   \(dp_{i+1,j}+dp_{i,j+1}=dp_{i,x}+dp_{x+1,j+1}+cost_{i,j+1}+dp_{i+1,y}+dp_{y+1,j}+cost_{i+1,j}\)
   很显然：\(dp_{i,j}+dp_{i+1,j+1} \le dp_{i+1,j}+dp_{i,j+1}\)

3. 令\(dp_{i,j}\)在\(best_{i,j}\)处取最优解，则\(best_{i,j-1} \le best_{i,j} \le best_{i+1,j}\)
   证明：令\(dp_{i,j-1}\)在\(k=y\)处取最优值，\(\forall x<y有x+1 \le y+1 \le j-1 < j\)
   \(dp_{x+1,j-1}+dp_{y+1,j} \le dp_{y+1,j}+dp_{x+1,j} \Rightarrow dp_{i,j-1}(k=x)-dp{i,j-1}(k=y) \le dp_{i,j}(k=x)+dp_{i,j}(k=y)\)
   故\(best_{i-1,j} \le best_{i,j} \le best_{i+1,j}\)

所以\(\forall dp_{i,j} = \max/\min\{dp_{i,k}+dp_{k+1,j}\}+cost_{i,j}(best_{i+1,j} \le k \le best_{i,j-1})\)

区间DP的时间复杂度就从\(O(n^3)\)降为\(O(n^2)\)