POJ 3264-Balanced Lineup（段树：单点更新，间隔查询）-白红宇

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34522		Accepted: 16224
Case Time Limit: 2000MS

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

Lines 2..

N+1: Line

i+1 contains a single integer that is the height of cow

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

Source

经ysj一说我也准备噜线段树了那天下午他来给我讲了一下线段树。先敲个模板再说。。

题意是找某个区间的最大值和最小值的差值。

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                 #define LL long longusing namespace std;const int INF=1<<27;const int maxn=200010;LL minn[maxn],maxx[maxn];void update(LL root,LL l,LL r,LL p,LL v)//单点更新{ if(l==r) maxx[root]=v;minn[root]=v; if(l
                 
                  =r) return minn[root]; if(ql<=mid) ans=min(ans,query_min(root*2,l,mid,ql,qr)); if(qr>mid) ans=min(ans,query_min(root*2+1,mid+1,r,ql,qr)); return ans;}LL query_max(LL root,LL l,LL r,LL ql,LL qr){ LL mid=(l+r)/2,ans=-INF; if(ql<=l&&qr>=r) return maxx[root]; if(ql<=mid) ans=max(ans,query_max(root*2,l,mid,ql,qr)); if(qr>mid) ans=max(ans,query_max(root*2+1,mid+1,r,ql,qr)); return ans;}int main(){ int N,Q,i,v; while(~scanf("%lld%lld",&N,&Q)) { for(i=1;i<=N;i++) { scanf("%lld",&v); update(1,1,N,i,v); } while(Q--) { int ql,qr; scanf("%lld%lld",&ql,&qr); printf("%lld\n",query_max(1,1,N,ql,qr)-query_min(1,1,N,ql,qr)); } } return 0;}

本文转自mfrbuaa博客园博客，原文链接：http://www.cnblogs.com/mfrbuaa/p/5041618.html，如需转载请自行联系原作者