专题二树形结构 E - Can you answer these queries V

1. 题目
   

   You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 and x1 <= x2 , y1 <= y2 }. Given M queries (1 <= M <= 10000), your program must output the results of these queries.

   Input

   The first line of the input consist of the number of tests cases <= 5. Each case consist of the integer N and the sequence A. Then the integer M. M lines follow, contains 4 numbers x1, y1, x2 y2.

   Output

   Your program should output the results of the M queries for each test case, one query per line.

   Example

   Input:
   2
   6 3 -2 1 -4 5 2
   2
   1 1 2 3
   1 3 2 5
   1 1
   1
   1 1 1 1
   
   Output:
   2
   3
   1
   
   
   

2. 思路
   这题没有修改，和A相比的区别主要是前后端点不定
   A题：https://www.cnblogs.com/Benincasa/p/15866592.html
   两个区间不相交时，最大连续和是x1，y1的最大后缀和，加上y1，x2的总和，再加上x2，y2的最大前缀和
   两个区间相交时有四种情况，一种情况下和不相交的时候方法一样，另外三种分别是x2，y1的最大连续和，x1，y1的最大连续和，x2，y2的最大连续和
   
3. 代码
   lmin是之前写别的思路的时候加的，没啥用
   

   #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, lmin, rsum, sum, maxx;
   }ss[M];
   
   int a[M << 2];
   int n,q,ca,ans;
   
   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].lmin = min(ss[p << 1].lmin, ss[p << 1].sum + ss[p << 1 | 1].lmin);
   	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].lmin = 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.lmin = min(ls.lmin, ls.sum + rs.lmin);
           ans.rsum = max(rs.rsum, rs.sum + ls.rsum);
           ans.sum = ls.sum + rs.sum;
           return ans;
       }
   }
   
   int main()
   {
   	scanf("%d",&ca);
   	while(ca--)
   	{
   		scanf("%d",&n);
   		for(int i=1;i<=n;i++){
   			scanf("%d",&a[i]);
   		}
   		build(1,n,1);
   		scanf("%d",&q);
   		int x1,y1,x2,y2;
   		while (q--) {
   			ans = 0;
   			cin >> x1 >> y1 >> x2 >> y2;
   			if(y1 < x2)
   			{
   				ans = query(x1,y1,1,n,1).rsum + query(x2,y2,1,n,1).lsum + query(y1,x2,1,n,1).sum;
   				ans = ans - a[y1] - a[x2];
   			}
   			else
   			{
   				int a1 = query(x2,y1,1,n,1).maxx;
   				int a2 = query(x1,x2,1,n,1).rsum + query(y1,y2,1,n,1).lsum + query(x2,y1,1,n,1).sum;
   				    a2 = a2 - a[y1] - a[x2];
   				int a3 = query(x1,x2,1,n,1).rsum + query(x2,y1,1,n,1).lsum - a[x2];
   				int a4 = query(x2,y1,1,n,1).rsum + query(y1,y2,1,n,1).lsum - a[y1];
   				ans = max(max(max(a1, a2), a3), a4);
   			}
   			cout << ans << endl;
   		}
   	}
   	return 0;
   }