answer:
very watery topic
First of all it's easy to see that each location actually only has the maximum value that is useful
Then change the condition to dp[i]=max(dp[j]+1)(j<i,F[i]>G[j],G[i]>H[j])
Then I researched a bit. . can be done on a rectangle
Then think again. . Isn't this a 3D partial order? .
So three-dimensional partial order can also be done by using a line segment tree over a line segment tree. . . Why didn't I know before