2022-05-26:void add(int L, int R, int C)代表在arrL...R上每个数加C,
int get(int L, int R)代表查询arrL...R上的累加和,
假设你可以在所有操作开始之前,重新排列arr。
请返回每一次get查询的结果都加在一起最大能是多少。
输入参数:
int[] arr : 原始数组,
int ops,二维数组每一行解释如下:
a,b,c,如果数组有3个数,表示调用add(a,b,c),
a,b,如果数组有2个数,表示调用get(a,b),
a和b表示arr范围,范围假设从1开始,不从0开始。
输出:
假设你可以在开始时重新排列arr,返回所有get操作返回值累计和最大是多少?
来自美团。
答案2022-05-26:
线段树。
代码用rust编写。代码如下:
fn main() {
let mut arr: Vec<i32> = vec![1, 2, 3, 4, 5];
let mut ops: Vec<Vec<i32>> = vec![vec![1, 3], vec![2, 4], vec![1, 5]];
println!("ans = {}", max_gets(&mut arr, &mut ops));
}
fn max_gets(arr: &mut Vec<i32>, ops: &mut Vec<Vec<i32>>) -> i32 {
let n = arr.len() as i32;
let mut get_tree = SegmentTree::new(n);
for op in ops.iter_mut() {
if op.len() == 2 {
get_tree.add(op[0], op[1], 1);
}
}
let mut get_cnts: Vec<Vec<i32>> = vec![];
for i in 0..n {
get_cnts.push(vec![]);
for _j in 0..2 {
get_cnts[i as usize].push(0);
}
}
let mut i: i32 = 1;
let mut j: i32 = 0;
while i <= n {
get_cnts[j as usize][0] = j;
get_cnts[j as usize][1] = get_tree.get(i, i);
i += 1;
j += 1;
}
get_cnts.sort_by(|a, b| a[1].cmp(&b[1]));
arr.sort();
let mut re_arrange: Vec<i32> = vec![];
for _i in 0..n {
re_arrange.push(0);
}
for i in 0..n {
re_arrange[get_cnts[i as usize][0] as usize] = arr[i as usize];
}
let mut st = SegmentTree::new2(&mut re_arrange);
let mut ans = 0;
for op in ops.iter_mut() {
if op.len() == 3 {
st.add(op[0], op[1], op[2]);
} else {
ans += st.get(op[0], op[1]);
}
}
return ans;
}
pub struct SegmentTree {
pub n: i32,
pub arr: Vec<i32>,
pub sum: Vec<i32>,
pub lazy: Vec<i32>,
}
impl SegmentTree {
pub fn new(size: i32) -> Self {
let mut n = size + 1;
let mut sum: Vec<i32> = vec![];
let mut lazy: Vec<i32> = vec![];
let arr: Vec<i32> = vec![];
for _i in 0..n << 2 {
sum.push(0);
lazy.push(0);
}
n -= 1;
Self { n, arr, sum, lazy }
}
pub fn new2(origin: &mut Vec<i32>) -> Self {
let mut n = origin.len() as i32 + 1;
let mut arr: Vec<i32> = vec![];
arr.push(0);
for i in 1..n {
arr.push(origin[(i - 1) as usize]);
}
let mut lazy: Vec<i32> = vec![];
let mut sum: Vec<i32> = vec![];
for _i in 0..n << 2 {
sum.push(0);
lazy.push(0);
}
n -= 1;
let mut a = Self { n, arr, sum, lazy };
a.build(1, n, 1);
return a;
}
fn build(&mut self, l: i32, r: i32, rt: i32) {
if l == r {
self.sum[rt as usize] = self.arr[l as usize];
return;
}
let mid = (l + r) >> 1;
self.build(l, mid, rt << 1);
self.build(mid + 1, r, rt << 1 | 1);
self.push_up(rt);
}
fn push_up(&mut self, rt: i32) {
self.sum[rt as usize] = self.sum[(rt << 1) as usize] + self.sum[(rt << 1 | 1) as usize];
}
fn push_down(&mut self, rt: i32, ln: i32, rn: i32) {
if self.lazy[rt as usize] != 0 {
self.lazy[(rt << 1) as usize] += self.lazy[rt as usize];
self.sum[(rt << 1) as usize] += self.lazy[rt as usize] * ln;
self.lazy[(rt << 1 | 1) as usize] += self.lazy[rt as usize];
self.sum[(rt << 1 | 1) as usize] += self.lazy[rt as usize] * rn;
self.lazy[rt as usize] = 0;
}
}
pub fn add(&mut self, ll: i32, rr: i32, cc: i32) {
self.add0(ll, rr, cc, 1, self.n, 1);
}
fn add0(&mut self, ll: i32, rr: i32, cc: i32, l: i32, r: i32, rt: i32) {
if ll <= l && r <= rr {
self.sum[rt as usize] += cc * (r - l + 1);
self.lazy[rt as usize] += cc;
return;
}
let mid = (l + r) >> 1;
self.push_down(rt, mid - l + 1, r - mid);
if ll <= mid {
self.add0(ll, rr, cc, l, mid, rt << 1);
}
if rr > mid {
self.add0(ll, rr, cc, mid + 1, r, rt << 1 | 1);
}
self.push_up(rt);
}
pub fn get(&mut self, ll: i32, rr: i32) -> i32 {
return self.query(ll, rr, 1, self.n, 1);
}
fn query(&mut self, ll: i32, rr: i32, l: i32, r: i32, rt: i32) -> i32 {
if ll <= l && r <= rr {
return self.sum[rt as usize];
}
let mid = (l + r) >> 1;
self.push_down(rt, mid - l + 1, r - mid);
let mut ans = 0;
if ll <= mid {
ans += self.query(ll, rr, l, mid, rt << 1);
}
if rr > mid {
ans += self.query(ll, rr, mid + 1, r, rt << 1 | 1);
}
return ans;
}
}
执行结果如下:
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。
原创声明:本文系作者授权腾讯云开发者社区发表,未经许可,不得转载。
如有侵权,请联系 cloudcommunity@tencent.com 删除。