Nyoj1185 maximum and minimum values (the maximum and minimum values of a line segment tree)

If you do not understand the line segment tree, read this article first. Click Open Link

The general method for this question, TLE.

Maximum and minimum time limit: 1000 MS | memory limit: 65535 kb difficulty: 2

Description

Give n integers and execute M queries. For each query, enter three integers C, L, and R:

If C is equal to 1, the minimum value between the number of L and the number of R is output;

If C is equal to 2, the maximum value between the number of L and the number of R is output;

If C is equal to 3, the sum of the minimum and maximum values between the number of L and the number of R is output.

(Including numbers L and R ).

Input

First, enter an integer T (T ≤ 100), indicating that there is a T group of data.
For each group of data, enter an integer N (1 ≤ n ≤ 10000) to indicate that there are n integers;
The next row contains N integers A (1 ≤ A ≤ 10000 );
Then input an integer m, which indicates that there are m inquiries;
Next there are m rows (1 ≤ m ≤ 10000), each row has 3 integers C, L, R (1 ≤ C ≤ 3, 1 ≤ L ≤ r ≤ n ).

Output

Output according to the description of the question. Each output occupies one row.

Sample Input

241 3 2 421 1 42 2 351 2 3 4 513 1 5

Sample output

136

Source

Original

#include <stdio.h>#include <algorithm>#define N 10005using namespace std;struct node {int left,right,min,max;}c[N*4];int a[N];void buildtree(int l,int r,int root){c[root].left=l;c[root].right=r;if(l==r){c[root].min=c[root].max=a[l];return ;}int mid=(l+r)/2;buildtree(l,mid,root*2);buildtree(mid+1,r,root*2+1);c[root].min=min(c[root*2].min,c[root*2+1].min);c[root].max=max(c[root*2].max,c[root*2+1].max);}void findtree(int l,int r,int root,int &min1,int &max1){if(c[root].left==l&&c[root].right==r){min1=c[root].min;max1=c[root].max;return ;}int mid=(c[root].left+c[root].right)/2;if(mid<l)findtree(l,r,root*2+1,min1,max1);else if(mid>=r)findtree(l,r,root*2,min1,max1);else{int min2,max2;findtree(l,mid,root*2,min1,max1);findtree(mid+1,r,root*2+1,min2,max2);    min1=min(min1,min2);max1=max(max1,max2);}}int main(){int t,n,x,l,r,sum,m,min1,max1;scanf("%d",&t);while(t--){scanf("%d",&n);for(int i=1;i<=n;i++)scanf("%d",&a[i]);buildtree(1,n,1);scanf("%d",&m);while(m--){scanf("%d %d %d",&x,&l,&r);findtree(l,r,1,min1,max1);if(x==1)printf("%d\n",min1);if(x==2)printf("%d\n",max1);if(x==3)printf("%d\n",min1+max1);}}return 0;}

Nyoj1185 maximum and minimum values (the maximum and minimum values of a line segment tree)