SP1716 GSS3 - Can you answer these queries III（线段树）

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{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

输入格式

The first line of input contains an integer N. The following line contains N integers, representing the sequence A1..AN.
The third line contains an integer M. The next M lines contain the operations in following form:
0 x y: modify Ax into y (|y|<=10000).
1 x y: print max{Ai + Ai+1 + .. + Aj | x<=i<=j<=y }.

输出格式

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

题意翻译

n 个数，q 次操作

操作0 x y把Ax 修改为y

操作1 l r询问区间[l,r] 的最大子段和

输入输出样例

输入 #1 复制

4

1 2 3 4

4

1 1 3

0 3 -3

1 2 4

1 3 3

输出 #1 复制

6

4

-3

 一切尽在注释。需要特别注意的是ask函数和别的题不太一样。

#include <bits/stdc++.h>
#define SIZE 50005
using namespace std;
int a[SIZE], n, q;
struct SegmentTree
{
    int p;
    int l;
    int r;
    int sum;
    int dat;
    int lmax;
    int rmax;
} t[4*SIZE];
void update(int p)
{
    t[p].sum = t[2 * p].sum + t[2 * p + 1].sum;
    t[p].lmax = max(t[2 * p].lmax, t[2 * p].sum + t[2 * p + 1].lmax);
    t[p].rmax = max(t[2 * p + 1].rmax, t[2 * p].rmax + t[2 * p + 1].sum);
    t[p].dat = max(max(t[2 * p].dat, t[2 * p + 1].dat), t[2 * p].rmax + t[2 * p + 1].lmax);
}
void build(int p, int l, int r)
{
    t[p].l = l, t[p].r = r;
    if(l == r)    {    t[p].dat = t[p].sum = t[p].lmax = t[p].rmax = a[l];    return;    }
    int mid = (l + r) >> 1;
    build(2 * p, l, mid);
    build(2 * p + 1, mid + 1, r);
    update(p);
}
void change(int p, int x, int v)
{
    if(t[p].l == t[p].r) {    t[p].dat = t[p].sum = t[p].lmax = t[p].rmax = v; return;    }//递归边界是当前左右端点相等 
    int mid = (t[p].l + t[p].r) >> 1;
    if(x <= mid) change(2 * p, x, v);//l<=mid !!!
    else change(2 * p + 1, x, v);
    update(p);
}
SegmentTree ask(int p, int l, int r)//注意ask的返回类型，这个题不是直接返回答案 
{
    if(l <= t[p].l && r >= t[p].r) return t[p];//包含在待询问区间内部 
    int mid = (t[p].l + t[p].r) >> 1;
    if(r <= mid) return ask(2 * p, l, r);//注意这里的参数还是l和r，因为待询问区间没有被mid分开 
    if(l > mid) return ask(2 * p + 1, l, r);
    //查询区间被mid分开了
    SegmentTree left = ask(2 * p, l, mid), right = ask(2 * p + 1, mid + 1, r), node;//对询问区间进行调整 在这里设置一个临时节点node，因为之前没有节点能同时统计到mid左右两边各一部分的信息 
    node.sum = left.sum + right.sum;//类似update函数 
    node.lmax = max(left.lmax, left.sum + right.lmax);
    node.rmax = max(right.rmax, left.rmax + right.sum);
    node.dat = max(max(left.dat, right.dat), left.rmax + right.lmax);
    return node;
}
int main()
{
    cin >> n;
    int i;
    for(i = 1; i <= n; i++) scanf("%d", &a[i]);
    build(1, 1, n);
    cin >> q;
    for(i = 1; i <= q; i++)
    {
        int c, a, b;
        scanf("%d%d%d", &c, &a, &b);
        if(c == 0) change(1, a, b);
        else cout << ask(1, a, b).dat<<endl;
    }
    return 0;
}