SPOJ GSS1 Can you answer these queries I

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.

线段树，每个节点维护区间和sum，从左端开始的最大值lmax，从右边开始的最大值rmax，和整段的最大值max。

由子段向上推的关系式如下。

sum=l.sum+r.sum

lmax=max(l.lmax,l.sum+r.lmax)  【前者不超过中点，后者贯穿中点】

rmax=max(r.rmax,r.sum+l.rmax)  【同上】

max=max(l.max,r.max,l.rmax+r.lmax)  【两边自己的最大值，或者两边连起来的最大值】

#include<cstdio>struct node{int lch,rch,l,r,lmax,rmax,max,sum;}t[200010];struct ans{int l,r,max,s;};int n,nn,a[50010];int mx(int x,int y){return x>y?x:y;}void build(int p,int l,int r){t[p].l=l;t[p].r=r;if (l==r){t[p].lmax=t[p].rmax=t[p].max=t[p].sum=a[l];return;}int mid=(t[p].l+t[p].r)/2;t[p].lch=++nn;build(nn,l,mid);t[p].rch=++nn;build(nn,mid+1,r);t[p].sum=t[t[p].lch].sum+t[t[p].rch].sum;t[p].lmax=mx(t[t[p].lch].lmax,t[t[p].lch].sum+t[t[p].rch].lmax);t[p].rmax=mx(t[t[p].rch].rmax,t[t[p].rch].sum+t[t[p].lch].rmax);t[p].max=mx(mx(t[t[p].lch].max,t[t[p].rch].max),t[t[p].lch].rmax+t[t[p].rch].lmax);}ans find(int p,int l,int r){ans al,ar,a;if (t[p].l==l&&t[p].r==r){a.l=t[p].lmax;a.r=t[p].rmax;a.max=t[p].max;a.s=t[p].sum;return a;}int mid=(t[p].l+t[p].r)/2;if (r<=mid) return find(t[p].lch,l,r);if (l>=mid+1) return find(t[p].rch,l,r);al=find(t[p].lch,l,mid);ar=find(t[p].rch,mid+1,r);a.l=mx(al.l,al.s+ar.l);a.r=mx(ar.r,ar.s+al.r);a.s=al.s+ar.s;a.max=mx(mx(al.max,ar.max),al.r+ar.l);return a;}int main(){int i,j,k,m,p,q,x,y,z;scanf("%d",&n);for (i=1;i<=n;i++)  scanf("%d",&a[i]);nn=1;build(1,1,n);scanf("%d",&m);for (i=1;i<=m;i++){scanf("%d%d",&x,&y);printf("%d\n",find(1,x,y).max);}}