51nod 1380 "jacket master of every three pumping a" (greedy + set)

 

Portal

 

• References

　　[. 1]: 51Nod-1380- jacket of every three pumping a master

• the meaning of problems

　　From the array of length n, extracted $ \ frac {n} {3} $ nonadjacent maximum value such that the sum (both can not be taken at the same time)

•answer

　　Select the current maximum greedy $ a_ {max} $, while deletion comprising $ a_ {max} $ including the left (A_L $ $), and the right thereof (A_R $ $) of these three digits;

　　However, it is also possible that $ a_l + a_r> a_ {max} $, then we need to choose $ a_l, a_r $ these two numbers, without selecting $ a_ {max} $ this number;

　　Therefore, we also need to $ a_l + a_r-a_ {max} $ into sequences to be considered;

　　Each time its left and its right to delete the number, may be two arrays $ L, R $ mark to record the current at its lower left and right positions thereof;

•Code

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 #define ll long long
 4 #define pll pair<ll ,ll >
 5 #define F first
 6 #define S second
 7 const int maxn=1e5+50;
 8 
 9 int n;
10 ll a[maxn];
11 int L[maxn];
12 int R[maxn];
13 set<pll >_set;
14 
15 void Del(int x)
16 {
17     _set.erase({a[x],x});
18     L[R[x]]=L[x];
19     R[L[x]]=R[x];
20 }
21 ll Solve()
22 {
23     for(int i=1;i <= n;++i)
24     {
25         L[i]=(i-1+n-1)%n+1;
26         R[i]=i%n+1;
27         _set.insert({a[i],i});
28     }
29 
30     ll ans=0;
31     for(int i=1;i <= n/3;++i)
32     {
33         ans += _set.rbegin()->F;
34 
35         int x=_set.rbegin()->S;
36         ll b=a[L[x]];
37         ll c=a[x];
38         ll d=a[R[x]];
39         
40         Del(L[x]);
41         Del(R[x]);
42         _set.erase({a[x],x});
43         
44         a[x]=b+d-c;
45         _set.insert({a[x],x});
46     }
47     return ans;
48 }
49 int main()
50 {
51     scanf("%d",&n);
52     for(int i=1;i <= n;++i)
53         scanf("%lld",a+i);
54     printf("%lld\n",Solve());
55 
56     return 0;
57 }

View Code