数据结构（线段树）：SPOJ GSS3 - Can you answer these queries III

GSS3 - Can you answer these queries III

You are given a sequence A of N (N <= 50000) integers between -10000 and 10000. On this sequence you have to apply M (M <= 50000) operations:
modify the i-th element in the sequence or for given x y print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Input

The first line of input contains an integer N. The following line contains N integers, representing the sequence A1..AN.
The third line contains an integer M. The next M lines contain the operations in following form:
0 x y: modify Ax into y (|y|<=10000).
1 x y: print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

Output

For each query, print an integer as the problem required.

Example

Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3
 
Output:
6
4
-3

 #include <iostream>
 #include <cstring>
 #include <cstdio>
 using namespace std;
 const int maxn=;
 const int INF=;
 int n,Q,a[maxn];
 
 struct Node{
     int lx,rx,mx,sum;
     Node(int _lx=-INF,int _rx=-INF,int _mx=-INF){
         lx=_lx;rx=_rx;mx=_mx;
     }
 }T[maxn<<];
 
 void Push_up(Node &x,Node l,Node r){
     x.sum=l.sum+r.sum;
     x.lx=max(l.lx,l.sum+r.lx);
     x.rx=max(r.rx,r.sum+l.rx);
     x.mx=max(max(l.mx,r.mx),l.rx+r.lx);
 }
 
 void Build(int x,int l,int r){
     if(l==r){
         T[x].lx=T[x].rx=max(a[l],);
         T[x].mx=T[x].sum=a[l];
         return;
     }
     int mid=(l+r)>>;
     Build(x<<,l,mid);
     Build(x<<|,mid+,r);
     Push_up(T[x],T[x<<],T[x<<|]);
 }
 
 void Modify(int x,int l,int r,int g){
     if(l==r){
         T[x].lx=T[x].rx=max(a[l],);
         T[x].mx=T[x].sum=a[l];
         return;
     }
     int mid=(l+r)>>;
     if(mid>=g)Modify(x<<,l,mid,g);
     else Modify(x<<|,mid+,r,g);
     Push_up(T[x],T[x<<],T[x<<|]);
 }
 
 Node Query(int x,int l,int r,int a,int b){
     if(l>=a&&r<=b)
         return T[x];
     int mid=(l+r)>>;
     Node L,R,ret;
     if(mid>=a)L=Query(x<<,l,mid,a,b);
     if(mid<b)R=Query(x<<|,mid+,r,a,b);
     Push_up(ret,L,R);
     return ret;
 }
 
 int main(){
     scanf("%d",&n);
     for(int i=;i<=n;i++)
         scanf("%d",&a[i]);
     Build(,,n);
     scanf("%d",&Q);
     int tp,x,y;
     while(Q--){
         scanf("%d%d%d",&tp,&x,&y);
         if(!tp)
             a[x]=y,Modify(,,n,x);
         else
             printf("%d\n",Query(,,n,x,y).mx);
     }
     return ;
 }

水题。