专题二树形结构 A - Can you answer these queries I

1. 题目
   

   You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
   Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y }.
   Given M queries, your program must output the results of these queries.

   Input

   • The first line of the input file contains the integer N.
   • In the second line, N numbers follow.
   • The third line contains the integer M.
   • M lines follow, where line i contains 2 numbers xi and yi.

   Output

   Your program should output the results of the M queries, one query per line.

   Example

   Input:
   3 
   -1 2 3
   1
   1 2
   
   Output:
   2
   

2. 思路
   题目要求给一组数，查询某区间内最大连续和。对于一个节点代表的区间，以及它的左儿子右儿子，有如下情况以及关系：
   （sum：区间和；lsum：区间最大前缀和；rsum：区间最大后缀和；maxx：区间最大连续元素的和）
   
   此题甚至没有修改区间的操作，摸清规律写线段树就行了
3. 代码
   这个实际上是C题代码，因为题目差别不大（C题只多了个修改），直接在A的基础上面改了，原来的找不到了，去掉change和输入的判断部分应该就是A的代码了
   

   #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;
   }