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You are given a sequence A[1], A[2], ..., A[N] . ( |A[i]| <= 10000 , 1 <= N <= 10000 ). A query is defined as follows: Query(x1,y1,x2,y2) = Max { A[i]+A[i+1]+...+A[j] ; x1 <= i <= y1 , x2 <= j <= y2 and x1 <= x2 , y1 <= y2 }. Given M queries (1 <= M <= 10000), your program must output the results of these queries.
The first line of the input consist of the number of tests cases <= 5. Each case consist of the integer N and the sequence A. Then the integer M. M lines follow, contains 4 numbers x1, y1, x2 y2.
Your program should output the results of the M queries for each test case, one query per line.
Input: 2 6 3 -2 1 -4 5 2 2 1 1 2 3 1 3 2 5 1 1 1 1 1 1 1 Output: 2 3 1
解题：线段树，两个区间交与不交 讨论即可
1 #include <bits/stdc++.h> 2 using namespace std; 3 const int maxn = 100010; 4 struct node{ 5 int lt,rt,lsum,rsum,sum,msum; 6 }tree[maxn<<2]; 7 int A[maxn]; 8 inline void pushup(node &a,node &b,node &c){ 9 c.sum = a.sum + b.sum; 10 c.lsum = max(a.lsum,a.sum + b.lsum); 11 c.rsum = max(b.rsum,b.sum + a.rsum); 12 c.msum = max(max(a.msum,b.msum),a.rsum + b.lsum); 13 } 14 void build(int lt,int rt,int v){ 15 tree[v].lt = lt; 16 tree[v].rt = rt; 17 if(lt == rt){ 18 scanf("%d",&tree[v].sum); 19 tree[v].lsum = tree[v].rsum = tree[v].msum = tree[v].sum; 20 A[lt] = tree[v].sum; 21 return; 22 } 23 int mid = (lt + rt)>>1; 24 build(lt,mid,v<<1); 25 build(mid + 1,rt,v<<1|1); 26 pushup(tree[v<<1],tree[v<<1|1],tree[v]); 27 } 28 node query(int lt,int rt,int v){ 29 if(lt > rt){ 30 node tmp; 31 tmp.sum = tmp.lsum = tmp.rsum = tmp.msum = 0; 32 return tmp; 33 } 34 if(lt <= tree[v].lt && rt >= tree[v].rt) return tree[v]; 35 if(rt <= tree[v<<1].rt) return query(lt,rt,v<<1); 36 if(lt >= tree[v<<1|1].lt) return query(lt,rt,v<<1|1); 37 node a = query(lt,rt,v<<1),b = query(lt,rt,v<<1|1),c; 38 pushup(a,b,c); 39 return c; 40 } 41 int main(){ 42 int kase,n,m,x1,y1,x2,y2; 43 scanf("%d",&kase); 44 while(kase--){ 45 scanf("%d",&n); 46 build(1,n,1); 47 scanf("%d",&m); 48 while(m--){ 49 scanf("%d%d%d%d",&x1,&y1,&x2,&y2); 50 if(y1 < x2){ 51 int a = query(x1,y1,1).rsum; 52 int b = query(x2,y2,1).lsum; 53 int c = x2 - y1 > 1?query(y1+1,x2-1,1).sum:0; 54 printf("%d\n",a + b + c); 55 }else{ 56 int a = query(x1,x2,1).rsum + query(x2,y1,1).lsum - A[x2]; 57 int b = query(x1,y1,1).rsum + query(y1,y2,1).lsum - A[y1]; 58 int c = query(x2,y1,1).msum; 59 printf("%d\n",max(max(a,b),c)); 60 } 61 } 62 } 63 return 0; 64 }
SPOJ GSS5 Can you answer these queries V
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原文地址：http://www.cnblogs.com/crackpotisback/p/4889858.html
