标签：
You are given a sequence $A[1], A[2], ..., A[N] $. $( |A[i]| \leq 15007 , 1\leq N \leq 50000 )$. A query is defined as follows:
$Query(x,y) = Max \{ a[i]+a[i+1]+...+a[j] ; x \leq i \leq j \leq y \}$.
Given M queries, your program must output the results of these queries.
Input: 3 -1 2 3 1 1 2 Output: 2
解题：线段树
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 50010; 4 struct node{ 5 int lt,rt,lsum,rsum,sum; 6 }tree[maxn<<2]; 7 inline void pushup(int v){ 8 tree[v].sum = tree[v<<1].sum + tree[v<<1|1].sum; 9 tree[v].lsum = max(tree[v<<1].lsum,tree[v<<1].sum + tree[v<<1|1].lsum); 10 tree[v].rsum = max(tree[v<<1|1].rsum,tree[v<<1|1].sum + tree[v<<1].rsum); 11 } 12 void build(int lt,int rt,int v){ 13 tree[v].lt = lt; 14 tree[v].rt = rt; 15 if(lt == rt){ 16 scanf("%d",&tree[v].sum); 17 tree[v].lsum = tree[v].rsum = tree[v].sum; 18 return; 19 } 20 int mid = (lt + rt)>>1; 21 build(lt,mid,v<<1); 22 build(mid + 1,rt,v<<1|1); 23 pushup(v); 24 } 25 node query(int lt,int rt,int v){ 26 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v]; 27 int mid = (tree[v].lt + tree[v].rt)>>1; 28 if(rt <= mid) return query(lt,rt,v<<1); 29 if(lt > mid) return query(lt,rt,v<<1|1); 30 node a = query(lt,rt,v<<1); 31 node b = query(lt,rt,v<<1|1); 32 node c; 33 c.sum = a.sum + b.sum; 34 c.lsum = max(a.lsum,a.sum + b.lsum); 35 c.rsum = max(b.rsum,b.sum + a.rsum); 36 return c; 37 } 38 int main(){ 39 int n,m,x,y; 40 while(~scanf("%d",&n)){ 41 build(1,n,1); 42 scanf("%d",&m); 43 while(m--){ 44 scanf("%d%d",&x,&y); 45 node d = query(x,y,1); 46 printf("%d\n",max(d.sum,max(d.lsum,d.rsum))); 47 } 48 } 49 return 0; 50 }
SPOJ GSS1 Can you answer these queries I
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原文地址：http://www.cnblogs.com/crackpotisback/p/4887164.html
