SPOJ KGSS - Maximum Sum

https://www.spoj.com/problems/KGSS/
You are given a sequence A[1], A[2], ..., A[N] ( 0 ≤ A[i] ≤ 10^8 , 2 ≤ N ≤ 10^5 ). There are two types of operations and they are defined as follows:

Update:

This will be indicated in the input by a 'U' followed by space and then two integers i and x.

U i x, 1 ≤ i ≤ N, and x, 0 ≤ x ≤ 10^8.

This operation sets the value of A[i] to x.

Query:

This will be indicated in the input by a 'Q' followed by a single space and then two integers i and j.

Q x y, 1 ≤ x < y ≤ N.

You must find i and j such that x ≤ i, j ≤ y and i != j, such that the sum A[i]+A[j] is maximized. Print the sum A[i]+A[j].

Input

The first line of input consists of an integer N representing the length of the sequence. Next line consists of N space separated integers A[i]. Next line contains an integer Q, Q ≤ 10^5, representing the number of operations. Next Q lines contain the operations.

Output

Output the maximum sum mentioned above, in a separate line, for each Query.

Example

Input:
5
1 2 3 4 5
6
Q 2 4
Q 2 5
U 1 6
Q 1 5
U 1 7
Q 1 5

Output:
7
9
11
12

https://kartikkukreja.wordpress.com/2014/11/09/a-simple-approach-to-segment-trees/

This problem asks for the maximum pair sum in each subarray and also requires updates to individual array elements. As it turns out, we only need to store 2 things in each segment tree node:

1. The maximum value in this range
2. The second maximum value in this range

https://gocodergo.wordpress.com/2017/06/14/spoj-kgss-maximum-sum/

Hint : This problem asks for finding the maximum sum of 2 numbers in a given range.
So the largest and second largest element in given range would give us the maximum sum.
Structure of a node in our segment tree :

1 2 3 4 5

struct tree{     int firstMax;      // largest element     int secondMax;     // second largest element     int sum;           // i used it in the following code but not required };

ll n;
const int N = 1e5 + 10;
ll a[ N << 2 ];
ll place[ N << 2 ];
struct tree{
    ll firstMax, secMax , sum;
};
tree t[ N << 2 ];
tree merge(tree foo, tree bar){
    vi v;
    v.pb(foo.firstMax);
    v.pb(bar.firstMax);
    v.pb(bar.secMax);
    v.pb(foo.secMax);
    sort(v.begin(), v.end());
    tree ret;
    ret.firstMax = v[3];
    ret.secMax = v[2];
    ret.sum = v[3] + v[2];
    return ret;
}


void constructSegmentTree(ll low, ll high, ll sgmnt){
    if(low == high){
        t[sgmnt].firstMax = a[low];
        t[sgmnt].secMax = 0;
        t[sgmnt].sum = a[low];
        place[low] = sgmnt;
        return;
    }

    int mid = low + (high - low) / 2;
    constructSegmentTree(low, mid, sgmnt*2+1);
    constructSegmentTree(mid+1, high, sgmnt*2+2);
    tree foo = t[sgmnt*2+1];
    tree bar = t[sgmnt*2+2];
    t[sgmnt] = merge(foo, bar);
}

void update(ll low, ll high, ll index, ll sgmnt){
    ll v = place[index];
    t[v].firstMax = t[v].sum = a[index];
    if(v & 1)
        v /= 2;
    else{
        v = v/2-1;
    }
    while(v >= 0){
        t[v] = merge(t[v * 2 + 1], t[v * 2 + 2]);
        if(v & 1)
           v /= 2;
        else{
            v = v/2-1;
        }
    }
}


tree query(ll low, ll high, ll x, ll y, ll sgmnt){
    if (x <= low && y >= high)
        return t[sgmnt];
    if(low > y || high < x){
        tree temp;
        temp.firstMax = temp.secMax = 0;
        temp.sum = 0;
        return temp;
    }
    int mid = low + (high - low) / 2;
    if(y <= mid){
        return query(low, mid, x, y, sgmnt * 2 + 1);
    }
    else if(x > mid){
        return query(mid+1, high, x, y, sgmnt * 2 + 2);
    }else{
        return merge(query(low, mid, x, y, sgmnt * 2 + 1), query(mid+1, high, x, y, sgmnt * 2 + 2));
    }
}

void solve(){
    cin >> n;
    ll i;
    rep(i,n) {
        cin >> a[i];
    }
    constructSegmentTree(0, n-1, 0);
    ll q;
    cin >> q;
    while(q--){
        char c;
        cin >> c;
        if(c == 'Q'){
            // Query
            ll x, y;
            cin >> x >> y;
            tree foo = query(0, n-1, x-1, y-1, 0);
            cout << foo.sum << endl;
        }else{
            // UPDATE A VALUE
            ll index, value;
            cin >> index >> value;
            index --;
            a[index] = value;
            update(0, n-1, index, 0);
        }
    }

}

https://blog.csdn.net/qq_36294146/article/details/79161490