Poj 3264 balanced lineup

Question link: http://poj.org/problem? Id = 3264

It is to give you a string of numbers and ask you the difference between the maximum number and the minimum number .......

Train of Thought: the most basic line segment tree only needs to build and query, and the modification is saved, but the query needs to write two. One query has the maximum value and the other one has the minimum value ...... Then the AC can be dropped ..... However, poj classifies it into rmq ....

Code:

# Include <cstdio> # include <cmath> # include <algorithm> # include <iostream> # define Max 50010 using namespace STD; struct node {int L, R; int Minn, maxx;} node [Max * 4]; int num [Max]; void buildtree (int I, int L, int R) {node [I]. L = L; node [I]. R = r; If (L = r) {node [I]. maxx = node [I]. minn = num [l]; return;} int mid = (L + r)/2; buildtree (I * 2, L, mid); buildtree (I * 2 + 1, mid + 1, R); node [I]. maxx = max (node [I * 2]. maxx, node [I * 2 + 1]. maxx); node [I]. minn = min (node [I * 2]. minn, node [I * 2 + 1]. minn);} int query (int I, int L, int R) // query the maximum value {If (node [I]. L = L & node [I]. R = r) {return node [I]. maxx;} int mid = (node [I]. L + node [I]. r)/2; If (r <= mid) return query (I * 2, L, R); else if (L> mid) return query (I * 2 + 1, l, R); else {return max (query (I * 2, L, mid), query (I * 2 + 1, Mid + 1, R ));}} int query1 (int I, int L, int R) // query the minimum value {If (node [I]. L = L & node [I]. R = r) {return node [I]. minn;} int mid = (node [I]. L + node [I]. r)/2; If (r <= mid) return query1 (I * 2, L, R); else if (L> mid) return query1 (I * 2 + 1, l, R); else {return (min (query1 (I * 2, L, mid), query1 (I * 2 + 1, Mid + 1, r) ;}} int main () {int N, Q; while (scanf ("% d", & N, & Q) = 2) {int I; for (I = 1; I <= N; I ++) {scanf ("% d", & num [I]) ;}buildtree (1, 1, n); for (I = 1; I <= Q; I ++) {int A, B; scanf ("% d", & A, & B ); int M1 = query (1, a, B); int m2 = query1 (1, a, B); printf ("% d \ n", M1-M2 );}} return 0 ;}