UVA 1471 Defense Lines line of defense

Given a sequence of length n, it is required to delete a continuous subsequence so that the remainder of the sequence has a continuous increment subsequence of the maximum length.

The simplest idea is to enumerate the starting J and the end point I, and then count the longest lengths that can be extended forward or backward, in the notation G (i) and f (i). The lengths at which each point can be stretched forward and backward can be preprocessed first (g (i) and f (i)). Then enumerate the end J, quickly find a G (j) the largest starting point. If there are two candidate starting points, a J, a J ', A[j ']<=a[j] and G[j ']>g[j], then J must be excluded, g (j ') larger, and easier to splice.

In the case of fixed I, all of the valuable (A[j],g (j)) sorted by A[j] (A[j] The same value retains the G (j) of the large one), will form an ordered table, according to the previous conclusion G (j) is incremental, orderly, then the two-point lookup is useful.

Then consider that the effect of the change I on the previous ordered table, I increase, the previous a[i],g (i) added to the ordered table, if satisfies A[i '] than it a[i] small and g (i ') the largest two-tuple, that is, one of its front elements, which satisfies g (i ') >=g (i), Then this element should not be preserved. Otherwise it should be added to this two-tuple, after adding this two-tuple, in order to maintain the nature of the ordered table, but also to check the deletion of some g (i*) small elements.

Finally think more thoroughly, the implementation of the way is set, with the pair to ensure the two-tuple, pair comparison is the first dimension, compared to the second dimension. As for the second implementation, use the array to save G (j) corresponding to the smallest a[j] value, review the LIS and then fill it ~

#include <bits/stdc++.h>#defineMP Make_pair#defineSe Second#defineFi firstusing namespacestd;Const intMAXN = 2e5+5; typedef pair<int,int>PII;intA[MAXN];intG[MAXN];//<-intF[MAXN];// -Set<pii>s;intMain () {intT scanf"%d",&T);  while(t--){        intN scanf"%d",&N);  for(inti =0; I < n; i++) scanf ("%d", A +i); if(n = =1) {printf ("1\n");Continue; } g[0] =1;  for(inti =1; I < n; i++) {            if(a[i]>a[i-1]) G[i] = g[i-1]+1; ElseG[i] =1; } f[n-1] =1;  for(inti = n2; I >=0; i--) {            if(a[i]<a[i+1]) F[i] = f[i+1]+1; ElseF[i] =1; } s.clear (); S.insert (MP (a[0],g[0])); intAns =1;  for(inti =1; I < n; i++) {PII cur (a[i],g[i]); Set<pii>::iterator it =s.lower_bound (cur); BOOLKeep =true; if(it!=S.begin ()) {It--; Ans= Max (ans,it->se+F[i]); Keep= It->se <cur.se; }            if(Keep) {s.erase (cur); S.insert (cur); It++; it++;  while(It! = S.end () && it->se <= cur.se) s.erase (it++); }} printf ("%d\n", ans); }    return 0;}

UVA 1471 Defense Lines line of defense