3304 fruit elder sister stroll Fruit Street Ⅰ
Time limit: 2 s
space limit: 256000 KB
title level: Diamonds Diamond
Title Description Description
Fruit sister today in a good mood, came to the fruit street.
Fruit Street has n fruit shop, in a straight line structure, numbered 1~n, each store can buy fruit can also sell fruit, and the same shop to sell and buy the same price.
Learn Oi fruit sister quickly found a way to make money: in a fruit shop to buy a fruit, and then sell to another shop, make a difference.
In the fruit when the elder sister, Cgh suddenly appeared, he in order to embarrass Fruit elder sister, give M a question, each question request fruit elder sister from the X home to the first shop, on the way only choose a shop to buy a fruit, then choose a shop (can be the same shop, but can't go back) sell out, Ask for the maximum amount of money you can make in each question.
Enter a description input Description
The first line is n, which means n stores
Down n A positive integer that represents the price of an apple per store.
Down an integer m, which represents down there m inquiries.
Down there are M-lines, two integers x and y per line, indicating the departure from the X home to the nth store.
outputs description output Description
There are M lines.
Each line corresponds to a query, an integer, indicating the face of cgh each query, fruit elder sister can earn up to how much money.
sample input to sample
2 8 1 (5) 3 5 4
6 6
2 8
2 2
6 3
sample output Sample outputs
0
0
14
Data Size & Hint
0<= Apple price <=10^8
0<n,m<=200000
The following:
The obvious line-segment tree. Maintain the interval maximum, minimum, the maximum difference from the front to the back, and the maximum difference from the backward forward. Just in the query when you need to discuss the classification, a bit disgusting.
Code:
#include <iostream> #include <cstdio> #include <cstdlib> #include <cstring> #include <cmath > #include <algorithm> #define root 1,1,n #define LCHILD rt<<1,l,m #define RCHILD rt<<1|1,m+1,r using
namespace Std; struct tree{int maxn,minn,chaq,chah;}
TREE[1000010];
int n,q;
int in () {int x=0; char Ch=getchar ();
while (ch< ' 0 ' | | ch> ' 9 ') Ch=getchar ();
while (ch>= ' 0 ' && ch<= ' 9 ') x=x*10+ch-' 0 ', Ch=getchar ();
return x;
} void push_up (int rt) {Tree[rt].maxn=max (TREE[RT<<1].MAXN,TREE[RT<<1|1].MAXN);
Tree[rt].minn=min (Tree[rt<<1].minn,tree[rt<<1|1].minn);
int q=tree[rt<<1].maxn-tree[rt<<1|1].minn,h=tree[rt<<1|1].maxn-tree[rt<<1].minn;
Tree[rt].chaq=max (Q,max (Tree[rt<<1].chaq,tree[rt<<1|1].chaq));
Tree[rt].chah=max (H,max (Tree[rt<<1].chah,tree[rt<<1|1].chah)); } void Build (int rt,int l,int R) {if (l==r) {TREE[RT].MAXN=tree[rt].minn=in ();
tree[rt].chaq=tree[rt].chah=0;
Return
} int m= (L+R) >>1; Build (Lchild);
Build (Rchild);
PUSH_UP (RT);
} tree Query (int rt,int l,int r,int ll,int rr) {if (ll<=l && r<=rr) return (Tree) TREE[RT]; int m= (L+R) >>1;
Tree S1,s2,s3;
S1=s2=s3= (Tree) {0,0x7fffffff,0,0};
if (ll<=m) s1= (Tree) query (LCHILD,LL,RR);
if (rr>m) s2= (Tree) query (RCHILD,LL,RR); S3.maxn=max (S1.MAXN,S2.MAXN);
S3.minn=min (S1.minn,s2.minn);
S3.chaq=max (Max (S1.chaq,s2.chaq), S1.maxn-s2.minn);
S3.chah=max (Max (S1.chah,s2.chah), S2.maxn-s1.minn);
return S3;
} int main () {n=in ();
Build (root);
Q=in ();
while (q--) {int x=in (), Y=in (), ans;
if (x==y) ans=0;
else if (x<y) ans=query (root,x,y). Chah;
else Ans=query (root,y,x). Chaq;
printf ("%d\n", ans);
} return 0; }