GSS3 - Can you answer these queries III

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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.

#include<bits/stdc++.h>#define MAXN 50010using namespace std;int lc[4*MAXN],rc[4*MAXN],mc[4*MAXN],sum[4*MAXN];int c[MAXN];void build(int id,int l,int r){if(l==r){sum[id]=mc[id]=lc[id]=rc[id]=c[l];return;}int mid=(l+r)/2;build(id*2,l,mid);build(id*2+1,mid+1,r);sum[id]=sum[id*2]+sum[id*2+1];lc[id]=max(lc[id*2],sum[id*2]+lc[id*2+1]);rc[id]=max(rc[id*2+1],sum[id*2+1]+rc[id*2]);mc[id]=max(max(mc[id*2],mc[id*2+1]),rc[id*2]+lc[id*2+1]);}int query(int id,int x,int y,int l,int r,int flag,int &ans){if(l<=x&&y<=r){ans=max(ans,mc[id]);return flag==-1?lc[id]:rc[id];}int mid=(x+y)/2;if(r<=mid)return query(id*2,x,mid,l,r,-1,ans);else if(l>mid)return query(id*2+1,mid+1,y,l,r,1,ans);else{int ln=query(id*2,x,mid,l,r,1,ans);int rn=query(id*2+1,mid+1,y,l,r,-1,ans);ans=max(ln+rn,ans);if(flag==-1)return max(lc[id*2],sum[id*2]+rn);elsereturn max(rc[id*2+1],sum[id*2+1]+ln);}}void update(int id,int l,int r,int pos,int val){if(l==r){sum[id]=lc[id]=rc[id]=mc[id]=val;return;}int mid=(l+r)/2;if(pos<=mid)update(id*2,l,mid,pos,val);elseupdate(id*2+1,mid+1,r,pos,val);sum[id]=sum[id*2]+sum[id*2+1];lc[id]=max(lc[id*2],sum[id*2]+lc[id*2+1]);rc[id]=max(rc[id*2+1],sum[id*2+1]+rc[id*2]);mc[id]=max(max(mc[id*2],mc[id*2+1]),rc[id*2]+lc[id*2+1]);}int main(){int n;scanf("%d",&n);    for(int i=1;i<=n;i++)    {        scanf("%d",&c[i]);    }   build(1,1,n);   int m;   scanf("%d",&m);   for(int i=0;i<m;i++)   {   int op,l,r;   scanf("%d%d%d",&op,&l,&r);   if(op==1)   {      int ans=c[l];   query(1,1,n,l,r,-1,ans);   printf("%d\n",ans);}else{c[l]=r;update(1,1,n,l,r);}}    return 0;}

在GSS1的基础上加入修改操作就行了；

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