POJ 3264 Balanced Lineup

题目链接

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

Sample Output

6
3
0

两种办法：
1.RMQ模板题，AC代码如下:

#include <cstdio>
#include<algorithm>
#include<cmath>
using namespace std;
typedef long long ll;
const int N=5e5+5;
int n,q,l,r,dp1[N][50],dp2[N][50],a[N];
void rmq_init()
{for(int i=1;i<=n;i++)dp1[i][0]=dp2[i][0]=a[i];for(int j=1;(1<<j)<=n;j++)for(int i=1;i+(1<<j)-1<=n;i++){dp1[i][j]=min(dp1[i][j-1],dp1[i+(1<<j-1)][j-1]);dp2[i][j]=max(dp2[i][j-1],dp2[i+(1<<j-1)][j-1]);}
}int rmqmin(int l,int r)
{int k=log2(r-l+1);return min(dp1[l][k],dp1[r-(1<<k)+1][k]);
}int rmqmax(int l,int r)
{int k=log2(r-l+1);return max(dp2[l][k],dp2[r-(1<<k)+1][k]);
}int main()
{scanf("%d%d",&n,&q);for(int i=1;i<=n;i++) scanf("%d",&a[i]);rmq_init();while(q--){scanf("%d%d",&l,&r);printf("%d\n",rmqmax(l,r)-rmqmin(l,r));}
}

2.线段树，消耗的内存小一些，时间一样

#include <cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N=5e4+5;
struct node{int mx,mn;
}tree[N<<2];
int n,q,l,r,mn,mx,a[N];
void pushup(int i)
{tree[i].mx=max(tree[i<<1].mx,tree[(i<<1)+1].mx);tree[i].mn=min(tree[i<<1].mn,tree[(i<<1)+1].mn);
}void build(int i,int l,int r){if(l==r){tree[i].mn=tree[i].mx=a[l];return ;}int mid=l+(r-l)/2;build(i<<1,l,mid);build((i<<1)+1,mid+1,r);pushup(i);
}void query(int i,int l,int r,int x,int y){if(x<=l&&r<=y){mx=max(mx,tree[i].mx);mn=min(mn,tree[i].mn);return ;}int mid=l+(r-l)/2;if(x<=mid) query(i<<1,l,mid,x,y);if(y>mid) query((i<<1)+1,mid+1,r,x,y);
}int main() {scanf("%d%d",&n,&q);for(int i=1;i<=n;i++) scanf("%d",&a[i]);build(1,1,n);while(q--){scanf("%d%d",&l,&r);mn=1e9,mx=-1e9;query(1,1,n,l,r);printf("%d\n",mx-mn);}return 0;
}