Balanced Lineup

Time Limit: 1 Sec Memory Limit: 256 MB

题目连接

http://poj.org/problem?id=3264

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 3

1 5

4 6

2 2

Sample Output

HINT

题意

给你n个数，然后查询区间的最大值减去最小值是多少

题解：

线段树RMQ，也可以ST来做，都是nlogn的复杂度

代码：

//qscqesze#include 
     
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                   typedef long long ll;using namespace std;//freopen("D.in","r",stdin);//freopen("D.out","w",stdout);#define sspeed ios_base::sync_with_stdio(0);cin.tie(0)#define maxn 200001#define mod 10007#define eps 1e-9int Num;char CH[20];//const int inf=0x7fffffff; //нчоч╢Сconst int inf=0x3f3f3f3f;/*inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}*///**************************************************************************************inline ll read(){ int x=0,f=1;char ch=getchar(); while(ch<'0'||ch>'9'){ if(ch=='-')f=-1;ch=getchar();} while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();} return x*f;}inline void P(int x){ Num=0;if(!x){putchar('0');puts("");return;} while(x>0)CH[++Num]=x%10,x/=10; while(Num)putchar(CH[Num--]+48); puts("");}struct node{ int l,r,ma,mi;}a[maxn];int d[maxn];void build(int x,int l,int r){ a[x].l=l,a[x].r=r; a[x].ma=-inf,a[x].mi=inf; if(l==r) { a[x].ma=d[l]; a[x].mi=d[l]; return; } else { int mid=(l+r)>>1; build(x<<1,l,mid); build(x<<1|1,mid+1,r); a[x].ma=max(a[x<<1].ma,a[x<<1|1].ma); a[x].mi=min(a[x<<1].mi,a[x<<1|1].mi); }}int query1(int x,int st,int ed){ int l=a[x].l,r=a[x].r; if(st<=l&&r<=ed) return a[x].mi; else { int mid=(l+r)>>1; int mi1=inf,mi2=inf; if(st<=mid) mi1=query1(x<<1,st,ed); if(ed>mid) mi2=query1(x<<1|1,st,ed); return min(mi1,mi2); }}int query2(int x,int st,int ed){ int l=a[x].l,r=a[x].r; if(st<=l&&r<=ed) return a[x].ma; else { int mid=(l+r)>>1; int mi1=-inf,mi2=-inf; if(st<=mid) mi1=query2(x<<1,st,ed); if(ed>mid) mi2=query2(x<<1|1,st,ed); return max(mi1,mi2); }}int main(){ int n=read(),q=read(); int dd,ee; for(int i=1;i<=n;i++) d[i]=read(); build(1,1,n); for(int i=0;i

转载于:https://www.cnblogs.com/qscqesze/p/4447979.html

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