专题二树形结构 C - Can you answer these queries III

1. 题目
   

   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
   

2. 思路
   A的基础上面加了个单点修改
   A题：https://www.cnblogs.com/Benincasa/p/15866592.html
3. 代码
   

   #include<cstdio>
   #include<iostream>
   #include<string>
   #include<cstring>
   #include<algorithm>
   using namespace std;
   
   #define clear(a,x) memset(a,x,sizeof(a))
   #define ll long long
   const int M=4e5;
   
   struct Node {
       int lsum, rsum, sum, maxx;
   }ss[M];
   
   int a[M << 2];//???
   //ll sum[M],lsum[M],rsum[M],maxx[M];
   int n,q;
   
   void up(int p)
   {
   	ss[p].sum = ss[p << 1].sum + ss[(p << 1) + 1].sum;
   	ss[p].lsum = max(ss[p << 1].lsum, ss[p << 1].sum + ss[p << 1 | 1].lsum);
   	ss[p].rsum = max(ss[p << 1 | 1].rsum, ss[p << 1 | 1].sum + ss[p << 1].rsum);
   	ss[p].maxx = max(max(ss[p << 1].maxx, ss[p << 1 | 1].maxx), ss[p << 1].rsum + ss[p << 1 | 1].lsum);
   }
   
   void build(int s, int t, int p) {
   
   	if (s == t) {
   	    ss[p].lsum = ss[p].rsum = ss[p].maxx = ss[p].sum = a[s];
   	    return;
   	}
   
   	int m = s + ((t - s) >> 1);
   	build(s, m, p << 1), build(m + 1, t, p << 1 | 1);
   	up(p);
   }
   
   void change(int x, int c, int s, int t, int p)
   {
   	int m = s + ((t - s) >> 1);
   	if (s == x && s == t) {
   		ss[p].maxx = ss[p].lsum = ss[p].rsum = ss[p].sum = c;
   		return;
   	}
   	if (x <= m) change(x, c, s, m, p << 1); else change(x, c, m+1, t, p << 1 | 1);
   	up(p);
   }
   
   Node query(int l, int r, int s, int t, int p){
       if (l <= s && t <= r) return ss[p];
       int m = s + ((t - s) >> 1);
       if (m >= r) return query(l, r, s, m, p << 1) ;
       else if (m < l) return query(l, r, m + 1, t, p << 1 | 1);
       else {
           Node ls = query(l, r, s, m, p << 1);
           Node rs = query(l, r, m + 1, t, p << 1 | 1);
           Node ans ;
           ans.maxx = max(max(ls.maxx, rs.maxx), ls.rsum + rs.lsum);
           ans.lsum = max(ls.lsum, ls.sum + rs.lsum);
           ans.rsum = max(rs.rsum, rs.sum + ls.rsum);
           ans.sum = ls.sum + rs.sum;
           return ans;
       }
   }
   
   int main()
   {
   	scanf("%d",&n);
   	for(int i=1;i<=n;i++){
   		scanf("%d",&a[i]);
   	}
   	build(1,n,1);
   	scanf("%d",&q);
   	int op,x,y;
   	while (q--) {
   		cin >> op >> x >> y;
   		if(op==1)
   		{
   			cout << query(x, y, 1, n, 1).maxx <<endl;
   		}
   		else
   		{
   			change(x, y, 1, n, 1);
   		}
   	}
   	return 0;
   }