3304 fruit elder sister stroll Fruit Street Ⅰ

3304 fruit elder sister stroll Fruit Street Ⅰ

time limit: 2 sspace limit: 256000 KBtitle level: Diamonds Diamond SolvingView Run ResultsTitle 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.

Output 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 Sample Input

10
2 8 15 1 10 5 19 19 3 5
4
6 6
2 8
2 2
6 3

Sample output Sample Output

0
18
0
14

Data range and Tips Data Size & Hint

0<= Apple Price <=10^8

0<n,m<=200000

Category labels Tags Click here to expandPartition-type DP dynamic programming interval DP segment tree tree structure

Exercises

For an interval, the answer is (take the example from left to right):

max{右区间的最大值-左区间的最小值，左区间的答案，右区间的答案}

Because it is possible to walk from right to left, the line tree records:

Maximum, minimum, answer from left to right, answer from right to left.

#include <bits/stdc++.h>using namespacestd;#defineN 200001#defineLC K<<1#defineRC K<<1|1#defineMid (L+r>>1)structss{intMaxn,minn,ans1,ans2;} A[n<<2];intXl[n],n,m;inlinevoidUpdata (intk) {A[K].MAXN=Max (A[LC].MAXN,A[RC].MAXN); A[k].minn=min (A[lc].minn,a[rc].minn); intt=a[rc].maxn-A[lc].minn; intc=a[lc].maxn-A[rc].minn; A[k].ans1=Max (T,max (a[lc].ans1,a[rc].ans1)); A[k].ans2=Max (C,max (A[RC].ANS2,A[LC].ANS2));}voidBuildintKintLintR) {    if(l==R) {A[K].MAXN=a[k].minn=Xl[l]; A[k].ans1=a[k].ans2=0; return;    } build (Lc,l,mid); Build (Rc,mid+1, R); Updata (k);} SS Query (intKintLintRintXinty) {    if(y<l| | X&GT;R)return(ss) {-1,100000001,-1,-1}; if(X&LT;=L&AMP;&AMP;Y&GT;=R)returnA[k]; SS Tmp1=query (Lc,l,mid,x,y), Tmp2=query (rc,mid+1, r,x,y), TMP; Tmp.ans1=max (Tmp1.ans1,max (tmp2.ans1,tmp2.maxn-Tmp1.minn)); Tmp.ans2=max (Tmp1.ans2,max (tmp2.ans2,tmp1.maxn-Tmp2.minn)); Tmp.minn=min (Tmp1.minn,tmp2.minn); TMP.MAXN=Max (TMP1.MAXN,TMP2.MAXN); returntmp;}intMain () {scanf ("%d",&N);  for(intI=1; i<=n;i++) scanf ("%d",&Xl[i]); Build (1,1, N); scanf ("%d",&m);  for(intI=1, l,r;i<=m;i++) {scanf ("%d%d",&l,&R); if(l<=r) printf ("%d\n", Query (1,1, N,l,r). ans1); Elseprintf ("%d\n", Query (1,1, n,r,l). Ans2); }    return 0;} 

3304 fruit elder sister stroll Fruit Street Ⅰ