思路很明确，直接用线段树.
| Balanced Lineup
Description For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height. Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group. Input Line 1: Two space-separated integers, N and Q. Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i Lines N+2.. N+ Q+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive. Output Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range. Sample Input 6 31734251 54 62 2 Sample Output 630 Source | |||||||||
#include#include #include using namespace std;#define N 50050#define INF 0x3fffffffint n,q;struct node{ int b,d; int mx,mi; node* Tl; node* Tr;}edge[10*N];int cnt;node *Tbuild(int s,int t){ node* tmp=&edge[cnt++]; tmp->mi=INF; tmp->mx=-1; tmp->Tl=tmp->Tr=NULL; tmp->b=s; tmp->d=t; int mid=(s+t)/2; if(s!=t) { tmp->Tl=Tbuild(s,mid); tmp->Tr=Tbuild(mid+1,t); } return tmp;}node* head;void insert(node* tmp,int s,int k){ if(s>tmp->mx) tmp->mx=s; if(s mi) tmp->mi=s; if(tmp->b==tmp->d&&tmp->d==k) return ; int mid = (tmp->b+tmp->d)/2; if(k<=mid) { insert(tmp->Tl,s,k); } else insert(tmp->Tr,s,k);}int findmx(node * tmp,int s,int t){ if(tmp->b==s&&tmp->d==t) { return tmp->mx; } int mid=(tmp->b+tmp->d)/2; if(s>mid) return findmx(tmp->Tr,max(mid+1,s),t); if(t<=mid) return findmx(tmp->Tl,s,min(mid,t)); return max(findmx(tmp->Tl,s,mid),findmx(tmp->Tr,mid+1,t));}int findmi(node * tmp,int s,int t){ if(tmp->b==s&&tmp->d==t) { return tmp->mi; } int mid=(tmp->b+tmp->d)/2; if(s>mid) return findmi(tmp->Tr,max(mid+1,s),t); if(t<=mid) return findmi(tmp->Tl,s,min(mid,t)); return min(findmi(tmp->Tl,s,mid),findmi(tmp->Tr,mid+1,t));}int main(){ cnt=0; scanf("%d%d",&n,&q); head=Tbuild(1,n); for(int i=1;i<=n;i++) { int x; scanf("%d",&x); insert(head,x,i); } for(int i=0;i