题解【SP1716】GSS3 - Can you answer these queries III

题目描述

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 $max\{A_i + A_{i+1} + .. + A_j$ $|$ $x<=i<=j<=y \}$.

输入输出格式

输入格式

The first line of input contains an integer $N$.

The following line contains $N$ integers, representing the sequence $A_{1}..A_{N}$.

The third line contains an integer $M$. The next $M$ lines contain the operations in following form:

0 x y: modify $A_x$ into $y$ $(|y|<=10000)$.

1 x y: print $max\{A_i + A_{i+1} + .. + A_j$ $|$ $x<=i<=j<=y\}$.

输出格式

For each query, print an integer as the problem required.

输入输出样例

输入样例#1

输出样例#1

题意翻译

$n$ 个数，$q$ 次操作

操作0 x y把$A_x$ 修改为$y$

操作1 l r询问区间$[l, r]$的最大子段和

题解

这题就是GSS1的带修改版本，建议先看一看我的题解，了解不带修改的版本怎么写。

本题的代码基于我GSS1的题解，一些注释也可以在那里看到。

这里讲一讲怎么修改：

与线段树相似，如果当前已经访问到了叶子节点，就直接将这个节点的所有参数都设为要修改的值即可，否则就递归左子树或右子树。最后记得上传节点！

下面给出$AC$代码：

#include <iostream>

#include <cstdio>

#include <cstdlib>

#include <cstring>

#include <algorithm>

#include <cmath>

#include <cctype>

using namespace std;

inline int gi()

{

    int f = 1, x = 0; char c = getchar();

    while (c < '0' || c > '9') { if (c == '-') f = -1; c = getchar();}

    while (c >= '0' && c <= '9') { x = x * 10 + c - '0'; c = getchar();}

    return f * x;

}

int n, m;

struct Node

{

	int sum, lans, rans, ans;

} t[50005 << 2];

inline void pushup(int x)

{

	t[x].sum = t[x << 1].sum + t[(x << 1) | 1].sum;

	t[x].lans = max(t[x << 1].lans, t[x << 1].sum + t[(x << 1) | 1].lans);

	t[x].rans = max(t[(x << 1) | 1].rans, t[(x << 1) | 1].sum + t[x << 1].rans);

	t[x].ans = max(max(t[x << 1].ans, t[(x << 1) | 1].ans), t[x << 1].rans + t[(x << 1) | 1].lans);

}

void bulid(int s, int o, int p)

{

	if (s == o)

	{

		t[p].sum = t[p].lans = t[p].rans = t[p].ans = gi();

		return;

	}

	int mid = (s + o) >> 1;

	bulid(s, mid, p << 1);

	bulid(mid + 1, o, (p << 1) | 1);

	pushup(p);

}

void modify(int l, int r, int s, int o, int p)//修改操作

{

	if (s == o)//已经是叶子节点了

	{

		t[p].ans = t[p].lans = t[p].rans = t[p].sum = r;//就更新它的所有参数

		return;

	}

	int mid = (s + o) >> 1;//找中点

	if (l <= mid) modify(l, r, s, mid, p << 1);//点在中点左边就递归左子树

	else modify(l, r, mid + 1, o, (p << 1) | 1);//否则递归右子树

	pushup(p);//上传节点

}

Node getans(int l, int r, int s, int o, int p)

{

	if (l <= s && r >= o)

	{

		return t[p];

	}

	int mid = (s + o) >> 1;

	if (l > mid) return getans(l, r, mid + 1, o, (p << 1) | 1);

	if (r <= mid) return getans(l, r, s, mid, p << 1);

	else

	{

		Node ans, a, b;

		a = getans(l, r, s, mid, p << 1), b = getans(l, r, mid + 1, o, (p << 1) | 1);

		ans.sum = a.sum + b.sum;

		ans.ans = max(max(a.ans, a.rans + b.lans), b.ans);

		ans.lans = max(a.lans, a.sum + b.lans);

		ans.rans = max(b.rans, b.sum + a.rans);

		return ans;

	}

}

int main()

{

	n = gi();

	bulid(1, n, 1);

	m = gi();

	for (int i = 1; i <= m; i++)

	{

		int fl = gi(), x = gi(), y = gi();

		if (fl == 1) printf("%d\n", getans(x, y, 1, n, 1).ans);//如果是要求答案就输出答案

		else modify(x, y, 1, n, 1);//否则就进行修改

	}

	return 0;

}