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

 1 #include <iostream>
 2 #include <cstring>
 3 #include <cstdio>
 4 using namespace std;
 5 const int maxn=50010;
 6 const int INF=1000000000;
 7 int n,Q,a[maxn];
 8
 9 struct Node{
10     int lx,rx,mx,sum;
11     Node(int _lx=-INF,int _rx=-INF,int _mx=-INF){
12         lx=_lx;rx=_rx;mx=_mx;
13     }
14 }T[maxn<<2];
15
16 void Push_up(Node &x,Node l,Node r){
17     x.sum=l.sum+r.sum;
18     x.lx=max(l.lx,l.sum+r.lx);
19     x.rx=max(r.rx,r.sum+l.rx);
20     x.mx=max(max(l.mx,r.mx),l.rx+r.lx);
21 }
22
23 void Build(int x,int l,int r){
24     if(l==r){
25         T[x].lx=T[x].rx=max(a[l],0);
26         T[x].mx=T[x].sum=a[l];
27         return;
28     }
29     int mid=(l+r)>>1;
30     Build(x<<1,l,mid);
31     Build(x<<1|1,mid+1,r);
32     Push_up(T[x],T[x<<1],T[x<<1|1]);
33 }
34
35 void Modify(int x,int l,int r,int g){
36     if(l==r){
37         T[x].lx=T[x].rx=max(a[l],0);
38         T[x].mx=T[x].sum=a[l];
39         return;
40     }
41     int mid=(l+r)>>1;
42     if(mid>=g)Modify(x<<1,l,mid,g);
43     else Modify(x<<1|1,mid+1,r,g);
44     Push_up(T[x],T[x<<1],T[x<<1|1]);
45 }
46
47 Node Query(int x,int l,int r,int a,int b){
48     if(l>=a&&r<=b)
49         return T[x];
50     int mid=(l+r)>>1;
51     Node L,R,ret;
52     if(mid>=a)L=Query(x<<1,l,mid,a,b);
53     if(mid<b)R=Query(x<<1|1,mid+1,r,a,b);
54     Push_up(ret,L,R);
55     return ret;
56 }
57
58 int main(){
59     scanf("%d",&n);
60     for(int i=1;i<=n;i++)
61         scanf("%d",&a[i]);
62     Build(1,1,n);
63     scanf("%d",&Q);
64     int tp,x,y;
65     while(Q--){
66         scanf("%d%d%d",&tp,&x,&y);
67         if(!tp)
68             a[x]=y,Modify(1,1,n,x);
69         else
70             printf("%d\n",Query(1,1,n,x,y).mx);
71     }
72     return 0;
73 }

水题。