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

2022-02-06 23:32:10 作者：互联网

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
```
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

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;
}

标签：y1,these,x2,int,queries,answer,query,y2,lsum
来源： https://www.cnblogs.com/Benincasa/p/15866626.html

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

Input

Output

Example