U93056 t2(树状数组+dp)
满分做法:
几组样例充分说明,最优的子序列单调区间最多不超过两个。所以可以用树状数组维护三个值,一个是单增的最大值,一个是单减的最大值,一个是有拐点的最大值。最后离散化倒叙处理即可
#include<cstring> #include<cmath> #include<iostream> #include<algorithm> #include<cstdio> using namespace std; typedef long long ll; const int maxm=1e5+7; int n; ll a[maxm],b[maxm]; ll c1[maxm]; ll c2[maxm]; ll c3[maxm]; int l; double ans=0; int lowbit(int x) { return x&(-x); } void add1(int x,ll y) { for(;x;x-=lowbit(x)) c1[x]=max(c1[x],y); } void add2(int x,ll y) { for(;x<=l;x+=lowbit(x)) c2[x]=max(c2[x],y); } void add3(int x,ll y) { for(;x;x-=lowbit(x)) c3[x]=max(c3[x],y); } ll ask1(int x) { ll ans=0; for(;x<=l;x+=lowbit(x)) ans=max(ans,c1[x]); return ans; } ll ask2(int x) { ll ans=0; for(;x;x-=lowbit(x)) ans=max(ans,c2[x]); return ans; } ll ask3(int x) { ll ans=0; for(;x<=l;x+=lowbit(x)) ans=max(ans,c3[x]); return ans; } int main() { scanf("%d",&n); for(int i=1;i<=n;i++) scanf("%lld",a+i),b[i]=a[i]; sort(b+1,b+n+1); l=unique(b+1,b+n+1)-b-1; for(int i=1;i<=n;i++) { a[i]=lower_bound(b+1,b+l+1,a[i])-b+1;//防止有0 } l++; for(int i=n;i>=1;i--) { ll w1=ask1(a[i]+1),w2=ask2(a[i]-1),w3=ask3(a[i]+1); ans=max(ans,(1.0*w3+b[a[i]-1])/2.0); ans=max(ans,(1.0*w2+b[a[i]-1])/2.0); ans=max(ans,1.0*w1+b[a[i]-1]); add1(a[i],b[a[i]-1]+w1); add2(a[i],b[a[i]-1]+w2); add3(a[i],b[a[i]-1]+max(w2,w3)); } printf("%.3lf",ans); return 0; }