线段树的核心是区间 与query(The core of segment tree is interval and query)-其他
线段树的核心是区间 与query(The core of segment tree is interval and query)
https://www.acwing.com/activity/content/code/content/167568/
需要查询的信息是 l,r区间里面的最大连续和
最大连续和由左右子节点l,r获得 =max(l最大字段和,r最大子段和,跨越中间的最大字段和(=左节点的右端最大字段和+右节点的左端最大字段和))
所以一个节点要存储的
l,r,lmax,rmax,ans
右因为 lmax的获得需要对应左右儿子节点的l.sum+r.left
所以还要维护一个sum
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 500010;
int n, m;
int w[N];
struct Node
{
int l, r;
int sum, lmax, rmax, tmax;
}tr[N * 4];
void pushup(Node &u, Node &l, Node &r)//直接传入本次要处理的节点和他的左右子节点 通过这样获得本次的节点信息
{
u.sum = l.sum + r.sum;
u.lmax = max(l.lmax, l.sum + r.lmax);
u.rmax = max(r.rmax, r.sum + l.rmax);
u.tmax = max(max(l.tmax, r.tmax), l.rmax + r.lmax);
}
void pushup(int u)
{
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void build(int u, int l, int r)
{
tr[u] = {l, r};
if (l == r)
{
tr[u] = {l, r, w[r], w[r], w[r], w[r]};
return ;
}
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
void modify(int u, int x, int v)
{
if (tr[u].l == x && tr[u].r == x){
tr[u] = {x, x, v, v, v, v};
return ;
}
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
Node query(int u, int l, int r)
{
if (tr[u].l >= l && tr[u].r <= r) return tr[u];
int mid = tr[u].l + tr[u].r >> 1;
if (r <= mid) return query(u << 1, l, r);
else if (l > mid) return query(u << 1 | 1, l, r);
else
{
auto left = query(u << 1, l, r);
auto right = query(u << 1 | 1, l, r);
Node res;
pushup(res, left, right);
return res;
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]);
build(1, 1, n);
int k, x, y;
while (m -- )
{
scanf("%d%d%d", &k, &x, &y);
if (k == 1)
{
if (x > y) swap(x, y);
printf("%d\n", query(1, x, y).tmax);//直接返回节点 然后输出节点的信息
}
else modify(1, x, y);
}
return 0;
}
————————
https://www.acwing.com/activity/content/code/content/167568/
The information to be queried is the maximum continuous sum in the L, R interval
The maximum continuous sum is obtained by the left and right sub nodes L, r = max (L maximum field sum, R maximum sub segment sum, the maximum field sum across the middle (= the maximum field sum at the right end of the left node + the maximum field sum at the left end of the right node))
So a node needs to store
l,r,lmax,rmax,ans
Because the right Lmax needs l.sum + r.left corresponding to the left and right son nodes
So you have to maintain a sum
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
const int N = 500010;
int n, m;
int w[N];
struct Node
{
int l, r;
int sum, lmax, rmax, tmax;
}tr[N * 4];
void pushup(Node &u, Node &l, Node &r)//直接传入本次要处理的节点和他的左右子节点 通过这样获得本次的节点信息
{
u.sum = l.sum + r.sum;
u.lmax = max(l.lmax, l.sum + r.lmax);
u.rmax = max(r.rmax, r.sum + l.rmax);
u.tmax = max(max(l.tmax, r.tmax), l.rmax + r.lmax);
}
void pushup(int u)
{
pushup(tr[u], tr[u << 1], tr[u << 1 | 1]);
}
void build(int u, int l, int r)
{
tr[u] = {l, r};
if (l == r)
{
tr[u] = {l, r, w[r], w[r], w[r], w[r]};
return ;
}
int mid = l + r >> 1;
build(u << 1, l, mid), build(u << 1 | 1, mid + 1, r);
pushup(u);
}
void modify(int u, int x, int v)
{
if (tr[u].l == x && tr[u].r == x){
tr[u] = {x, x, v, v, v, v};
return ;
}
int mid = tr[u].l + tr[u].r >> 1;
if (x <= mid) modify(u << 1, x, v);
else modify(u << 1 | 1, x, v);
pushup(u);
}
Node query(int u, int l, int r)
{
if (tr[u].l >= l && tr[u].r <= r) return tr[u];
int mid = tr[u].l + tr[u].r >> 1;
if (r <= mid) return query(u << 1, l, r);
else if (l > mid) return query(u << 1 | 1, l, r);
else
{
auto left = query(u << 1, l, r);
auto right = query(u << 1 | 1, l, r);
Node res;
pushup(res, left, right);
return res;
}
}
int main()
{
scanf("%d%d", &n, &m);
for (int i = 1; i <= n; i ++ ) scanf("%d", &w[i]);
build(1, 1, n);
int k, x, y;
while (m -- )
{
scanf("%d%d%d", &k, &x, &y);
if (k == 1)
{
if (x > y) swap(x, y);
printf("%d\n", query(1, x, y).tmax);//直接返回节点 然后输出节点的信息
}
else modify(1, x, y);
}
return 0;
}