SPOJ GSS3 Can you answer these queries III

 

Time Limit: 330MS   Memory Limit: 1572864KB   64bit IO Format: %lld & %llu

Description

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

Hint

Added by: Bin Jin
Date: 2007-08-03
Time limit: 0.330s
Source limit: 5000B
Memory limit: 1536MB
Cluster: Cube (Intel G860)
Languages: All except: C++ 5
Resource: own problem

 

单点修改，询问区间内最大连续字段和。

@TYVJ P1427 小白逛公园

 1 /*by SilverN*/
 2 #include<algorithm>
 3 #include<iostream>
 4 #include<cstring>
 5 #include<cstdio>
 6 #include<cmath>
 7 #define lc rt<<1
 8 #define rc rt<<1|1
 9 using namespace std;
10 const int mxn=100010;
11 int read(){
12     int x=0,f=1;char ch=getchar();
13     while(ch<'0' || ch>'9'){if(ch=='-')f=-1;ch=getchar();}
14     while(ch>='0' && ch<='9'){x=x*10+ch-'0';ch=getchar();}
15     return x*f;
16 }
17 int n,m;
18 int data[mxn];
19 struct node{
20     int mx;
21     int ml,mr;
22     int smm;
23 }t[mxn<<2],tmp0;
24 void push_up(int l,int r,int rt){
25     t[rt].smm=t[lc].smm+t[rc].smm;
26     t[rt].mx=max(t[lc].mx,t[rc].mx);
27     t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);
28     t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);
29     t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);
30     return;
31 }
32 void Build(int l,int r,int rt){
33     if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}
34     int mid=(l+r)>>1;
35     Build(l,mid,lc);
36     Build(mid+1,r,rc);
37     push_up(l,r,rt);
38     return;
39 }
40 void change(int p,int v,int l,int r,int rt){
41     if(l==r){
42         if(p==l){t[rt].ml=t[rt].mr=t[rt].mx=t[rt].smm=v;}
43         return;
44     }
45     int mid=(l+r)>>1;
46     if(p<=mid)change(p,v,l,mid,lc);
47     else change(p,v,mid+1,r,rc);
48     push_up(l,r,rt);
49     return;
50 }
51 node query(int L,int R,int l,int r,int rt){
52 //    printf("%d %d %d %d %d\n",L,R,l,r,rt);
53     if(L<=l && r<=R){return t[rt];}
54     int mid=(l+r)>>1;
55     node res1;
56     if(L<=mid)res1=query(L,R,l,mid,lc);
57         else res1=tmp0;
58     node res2;
59     if(R>mid)res2=query(L,R,mid+1,r,rc);
60         else res2=tmp0;
61     node res={0};
62     res.smm=res1.smm+res2.smm;
63     res.mx=max(res1.mx,res2.mx);
64     res.mx=max(res.mx,res1.mr+res2.ml);
65     res.ml=max(res1.ml,res1.smm+res2.ml);
66     res.mr=max(res2.mr,res2.smm+res1.mr);
67     return res;
68 }
69 int main(){
70     n=read();
71     int i,j,x,y,k;
72     for(i=1;i<=n;i++)data[i]=read();
73     Build(1,n,1);
74     tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0;
75     m=read();
76     for(i=1;i<=m;i++){
77         k=read();x=read();y=read();
78         if(k){
79             if(x>y)swap(x,y);
80             printf("%d\n",query(x,y,1,n,1).mx);
81         }
82         else change(x,y,1,n,1);
83     }
84     return 0;
85 }

 

posted @ 2016-11-30 00:15  SilverNebula  阅读(382)  评论(0)  编辑  收藏  举报
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