SP1716 GSS3

题意翻译

nn 个数，qq 次操作

操作0 x y把A_xAx​ 修改为yy

操作1 l r询问区间[l, r][l,r] 的最大子段和

感谢 @Edgration 提供的翻译

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 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.

输入输出样例

输入样例#1： 复制

4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

输出样例#1： 复制

6
4
-3

其实很基础的一道线段树，但是赛场上就是调不通，又是一次赛后AC的典范，一定是我写代码的方式不对。。。

#include <bits/stdc++.h>
#define mid ((t[d].l + t[d].r) >> 1)
using namespace std;
const int N = 50005;
struct Tree{
    int l,r,lgss,rgss,gss,sum;
}t[N*8];

int n,m;
int a[N];

void build(int d,int l,int r){
    t[d].l = l;t[d].r = r;
    if (l == r){
        t[d].gss = t[d].lgss = t[d].rgss = t[d].sum = a[l];
        return;
    }
    build(d*2,l,mid);
    build(d*2+1,mid+1,r);
    t[d].sum = t[d*2].sum + t[d*2+1].sum;
    t[d].gss = max(t[d*2].gss,max(t[d*2+1].gss,t[d*2].rgss+t[d*2+1].lgss));
    t[d].lgss = max(t[d*2].lgss,t[d*2].sum + t[d*2+1].lgss);
    t[d].rgss = max(t[d*2+1].rgss,t[d*2].rgss + t[d*2+1].sum);
}

void modify(int d,int x,int y){
    if (t[d].l == x && t[d].r == x){
        t[d].gss = t[d].lgss = t[d].rgss = t[d].sum = y;
        return;
    }
    if (x <= mid) modify(d*2,x,y);
    else if (x > mid) modify(d*2+1,x,y);
    t[d].sum = t[d*2].sum + t[d*2+1].sum;
    t[d].gss = max(t[d*2].gss,max(t[d*2+1].gss,t[d*2].rgss+t[d*2+1].lgss));
    t[d].lgss = max(t[d*2].lgss,t[d*2].sum + t[d*2+1].lgss);
    t[d].rgss = max(t[d*2+1].rgss,t[d*2].rgss + t[d*2+1].sum);
}

int getl(int d,int l,int r){
    if (t[d].l == l && t[d].r == r){
        return t[d].lgss;
    }
    if (r > mid){
        return max(getl(d*2,l,mid),t[d*2].sum + getl(d*2+1,mid+1,r));
    }
    return getl(d*2,l,r);
}

int getr(int d,int l,int r){
    if (t[d].l == l && t[d].r == r){
        return t[d].rgss;
    }
    if (l <= mid){
        return max(getr(d*2+1,mid+1,r),getr(d*2,l,mid) + t[d*2+1].sum);
    }else return getr(d*2+1,l,r);
}

int get(int d,int l,int r){
    if (t[d].l == l && t[d].r == r){
        return t[d].gss;
    }
    if (r <= mid) return get(d*2,l,r);
    else if (l > mid) return get(d*2+1,l,r);
    return max(get(d*2,l,mid),max(get(d*2+1,mid+1,r),getr(d*2,l,mid)+getl(d*2+1,mid+1,r)));
}

int main(){
    scanf("%d",&n);
    for (int i = 1;i <= n;++i){
        scanf("%d",a+i);

    }
    build(1,1,n);
    scanf("%d",&m);
    for (int i = 0;i < m;++i){
        int kk,x,y;
        scanf("%d%d%d",&kk,&x,&y);
        if (kk == 0){
            modify(1,x,y);
        }else {
            printf("%d
",get(1,x,y));
        }
    }
}