Poj 3264 balanced lineup

The question is: a farmer has a nheaded ox, and the height of each ox is different, so that you can find out the height difference between the highest ox and the lowest ox in the specified range.

Solution:

When I saw the question, I thought of the Line Segment tree. Then I used the recursive line segment tree. At the beginning, I had to replace it with a non-recursive version. Then I optimized the input and output for, this made me feel the beauty of optimization !!!

 1 #include<iostream> 2 #include<string> 3 #include<algorithm> 4 #define maxn 50100 5  6 using namespace std; 7  8 int MAX[maxn<<2], MIN[maxn<<2]; 9 int a[maxn];10 11 void PushUp(int rt)12 {13     MAX[rt]=max(MAX[rt<<1],MAX[rt<<1|1]);14     MIN[rt]=min(MIN[rt<<1],MIN[rt<<1|1]);15 }16 17 void Build(int l,int r,int rt)18 {19     if(l==r)20     {21         MAX[rt]=a[l];22         MIN[rt]=a[l];23     }24     else25     {26         int m=(l+r)>>1;27         Build(l,m,rt<<1);28         Build(m+1,r,rt<<1|1);29         PushUp(rt);30     }31 }32 33 void Query(int L,int R,int l,int r,int rt, int &mx, int &mn)34 {35     if(L<=l&&r<=R)36     {37         mx = max(MAX[rt], mx);38         mn = min(MIN[rt], mn);39         return;40     }41     int m=(r+l)>>1;42     if(L<=m)43         Query(L,R,l,m,rt<<1,mx,mn);44     if(R>m)45         Query(L,R,m+1,r,rt<<1|1,mx,mn);46 }47 48 49 int main()50 {51     int n, q;52     scanf("%d%d", &n, &q);53     for(int i=1; i<=n; ++i)54         scanf("%d",&a[i]);55     Build(1,n,1);56     int c, d, mx, mn;57     for(int i=0; i<q; ++i)58     {59         mx = 0, mn = 10000000;60         scanf("%d %d", &c, &d);61         Query(c,d,1,n,1, mx, mn);62         printf("%d\n", mx-mn); 63     }64     return 0;65 }

For non-recursive versions, see this article: 48299459.

Poj 3264 balanced lineup