Can you answer these queries III
Time Limit: 2000ms
Memory Limit: 262144KB
This problem will be judged on SPOJ. Original ID: GSS364-bit integer IO format: %lld Java class name: Main
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
Source
解题：带修改的最大字段和
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 50010; 4 struct node { 5 int lt,rt,lsum,rsum,sum,msum; 6 } tree[maxn<<2]; 7 inline void pushup(int v) { 8 tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum; 9 tree[v].lsum = max(tree[v<<1].lsum,tree[v<<1].sum + tree[v<<1|1].lsum); 10 tree[v].rsum = max(tree[v<<1|1].rsum,tree[v<<1|1].sum + tree[v<<1].rsum); 11 tree[v].msum = max(max(tree[v<<1].msum,tree[v<<1|1].msum),tree[v<<1].rsum + tree[v<<1|1].lsum); 12 } 13 void build(int lt,int rt,int v) { 14 tree[v].lt = lt; 15 tree[v].rt = rt; 16 if(lt == rt) { 17 scanf("%d",&tree[v].sum); 18 tree[v].msum = tree[v].lsum = tree[v].rsum = tree[v].sum; 19 return; 20 } 21 int mid = (lt + rt)>>1; 22 build(lt,mid,v<<1); 23 build(mid + 1,rt,v<<1|1); 24 pushup(v); 25 } 26 void update(int pos,int val,int v) { 27 if(tree[v].lt == tree[v].rt) { 28 tree[v].sum = (tree[v].rt - tree[v].lt + 1)*val; 29 tree[v].lsum = tree[v].rsum = tree[v].msum = max(tree[v].sum,val); 30 return; 31 } 32 if(pos <= tree[v<<1].rt) update(pos,val,v<<1); 33 if(pos >= tree[v<<1|1].lt) update(pos,val,v<<1|1); 34 pushup(v); 35 } 36 node query(int lt,int rt,int v) { 37 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v]; 38 int mid = (tree[v].lt + tree[v].rt)>>1; 39 if(rt <= mid) return query(lt,rt,v<<1); 40 if(lt > mid) return query(lt,rt,v<<1|1); 41 node a = query(lt,rt,v<<1); 42 node b = query(lt,rt,v<<1|1); 43 node c; 44 c.sum = a.sum + b.sum; 45 c.lsum = max(a.lsum,a.sum + b.lsum); 46 c.rsum = max(b.rsum,b.sum + a.rsum); 47 c.msum = max(max(a.msum,b.msum),a.rsum + b.lsum); 48 return c; 49 } 50 int main() { 51 int n,m,x,y,op; 52 while(~scanf("%d",&n)) { 53 build(1,n,1); 54 scanf("%d",&m); 55 while(m--) { 56 scanf("%d%d%d",&op,&x,&y); 57 if(op) { 58 node d = query(x,y,1); 59 printf("%d ",max(max(d.sum,d.msum),max(d.lsum,d.rsum))); 60 } else update(x,y,1); 61 } 62 } 63 return 0; 64 }