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SPOJ GSS1 Can you answer these queries I
Time Limit: 115MS | Memory Limit: 1572864KB | 64bit IO Format: %lld & %llu |
Description
You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| ≤ 15007 , 1 ≤ N ≤ 50000 ). A query is defined as follows:
Query(x,y) = Max { a[i]+a[i+1]+...+a[j] ; x ≤ i ≤ j ≤ y }.
Given M queries, your program must output the results of these queries.
Input
- The first line of the input file contains the integer N.
- In the second line, N numbers follow.
- The third line contains the integer M.
- M lines follow, where line i contains 2 numbers xi and yi.
Output
- Your program should output the results of the M queries, one query per line.
Example
Input:3 -1 2 311 2Output:2
Hint
Added by: | Nguyen Dinh Tu |
Date: | 2006-11-01 |
Time limit: | 0.115s-0.230s |
Source limit: | 5000B |
Memory limit: | 1536MB |
Cluster: | Cube (Intel G860) |
Languages: | All except: ERL JS NODEJS PERL 6 VB.net |
询问区间内和最大的连续子序列。没有修改操作。
线段树维护即可。
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 Build(int l,int r,int rt){25 if(l==r){t[rt].mx=t[rt].ml=t[rt].mr=data[l];t[rt].smm=data[l];return;}26 int mid=(l+r)>>1;27 Build(l,mid,lc);28 Build(mid+1,r,rc);29 t[rt].smm=t[lc].smm+t[rc].smm;30 t[rt].mx=max(t[lc].mx,t[rc].mx);31 t[rt].mx=max(t[lc].mr+t[rc].ml,t[rt].mx);32 t[rt].ml=max(t[lc].ml,t[lc].smm+t[rc].ml);33 t[rt].mr=max(t[rc].mr,t[rc].smm+t[lc].mr);34 return;35 }36 node query(int L,int R,int l,int r,int rt){37 // printf("%d %d %d %d %d\n",L,R,l,r,rt);38 if(L<=l && r<=R){return t[rt];}39 int mid=(l+r)>>1;40 node res1;41 if(L<=mid)res1=query(L,R,l,mid,lc);42 else res1=tmp0;43 node res2;44 if(R>mid)res2=query(L,R,mid+1,r,rc);45 else res2=tmp0;46 node res={0};47 res.smm=res1.smm+res2.smm;48 res.mx=max(res1.mx,res2.mx);49 res.mx=max(res.mx,res1.mr+res2.ml);50 res.ml=max(res1.ml,res1.smm+res2.ml);51 res.mr=max(res2.mr,res2.smm+res1.mr);52 return res;53 }54 int main(){55 n=read();56 int i,j,x,y;57 for(i=1;i<=n;i++)data[i]=read();58 Build(1,n,1);59 m=read();60 tmp0.ml=tmp0.mr=tmp0.mx=-1e9;tmp0.smm=0;61 for(i=1;i<=m;i++){62 x=read();y=read();63 printf("%d\n",query(x,y,1,n,1).mx);64 }65 return 0;66 }
SPOJ GSS1 Can you answer these queries I
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