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最新推荐文章于 2021-09-07 23:37:34 发布

zuihoudebingwen

最新推荐文章于 2021-09-07 23:37:34 发布

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分类专栏：线段树文章标签： tree integer build query struct output

本文链接：https://blog.youkuaiyun.com/zuihoudebingwen/article/details/8018426

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本文介绍了一种处理操作序列的问题，包括修改元素值和查询最大子序列和。使用线段树数据结构实现高效的查询与修改操作。

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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 }.

Input

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 }.

Output

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

Example

Input:
4
1 2 3 4
4
1 1 3
0 3 -3
1 2 4
1 3 3

Output:
6
4
-3

题意
维护一个数列a[1], a[2], ..., a[N] . (|a[i]| ≤ 10000, 1 ≤ N ≤ 50000)。
有两种共M个操作（M <= 50000）：Modify(i, x) 把a[i]改成x；Query(x, y) = max{a[i] + a[i + 1] + ... + a[j]; x ≤ i ≤ j ≤ y}。

题解
只比GSS1多了修改。

#include<iostream>
#include<cstdio>
#include<cmath>
#define maxn 50005
#define inf 1000000000
#define lc root<<1
#define rc root<<1|1
using namespace std;
int a[maxn];
   struct tree
{
    int maxs,maxl,maxr,alls;
    tree(int x=0)
    {
        maxs=maxl=maxr=alls=x;
    }
}data[maxn<<2];
tree update(tree a,tree b)
{
    tree ret;
    ret.alls=a.alls+b.alls;
    ret.maxs=max(a.maxr+b.maxl,max(a.maxs,b.maxs));
    ret.maxl=max(a.maxl,a.alls+b.maxl);
    ret.maxr=max(b.maxr,b.alls+a.maxr);
    return ret;
}
void build(int root,int l,int r)
{
    if(l==r)
    {
        data[root]=tree(a[l]);return;
    }
    int mid=(l+r)>>1;
    build(lc,l,mid);
    build(rc,mid+1,r);
    data[root]=update(data[lc],data[rc]);
}
void modify(int root,int l,int r,int id ,int val)
{
    if(l==r)
    {
        data[root]=tree(val);return;
    }
    int mid=(l+r)>>1;
    if(id<=mid)
    modify(lc,l,mid,id,val);
    else
    modify(rc,mid+1,r,id,val);
    data[root]=update(data[lc],data[rc]);
}
tree find(int root,int aa,int bb,int l,int r)
{
    if(l<=aa&&r>=bb)
    return data[root];
    tree x=tree(-inf),y=tree(-inf);
    x.alls=0;y.alls=0;
    int mid=(aa+bb)>>1;
    if(l<=mid)
    x=find(lc,aa,mid,l,r);
    if(r>mid)
    y=find(rc,mid+1,bb,l,r);
    return update(x,y);
}


int main()
{
    int n,m,i,j,x,y,f;
    while(scanf("%d",&n)!=EOF)
    {
        for(i=1;i<=n;i++)
        scanf("%d",&a[i]);
        build(1,1,n);
        scanf("%d",&m);
        while(m--)
        {
            scanf("%d%d%d",&f,&x,&y);
            if(f==1)
            printf("%d\n",find(1,1,n,x,y).maxs);
            else
            {
                modify(1,1,n,x,y);
            }
        }
    }
    return 0;
}