poj 3264 Balanced Lineup rmq

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 3
1
7
3
4
2
5
1 5
4 6
2 2

Sample Output

6
3
0

Source

USACO 2007 January Silver

思路：rmq 板子题，区间最大值-区间最小值，无更新；

#include<iostream>
#include<cstdio>
#include<cmath>
#include<string>
#include<queue>
#include<algorithm>
#include<stack>
#include<cstring>
#include<vector>
#include<list>
#include<set>
#include<map>
using namespace std;
#define ll long long
//#define mod 1000000007
#define pi (4*atan(1.0))
const int N=1e5+10,M=1e6+10,inf=1e9+10;
int a[N];
int dpi[N][30];//存位置
int dpa[N][30];
int minn(int x,int y)
{
    return a[x]<=a[y]?x:y;
}
void rmqi(int len)
{
    for(int i=0; i<len; i++)
    dpi[i][0]=i;
    for(int j=1; (1<<j)<len; j++)
    for(int i=0; i+(1<<j)-1<len; i++)
    dpi[i][j]=minn(dpi[i][j-1],dpi[i+(1<<(j-1))][j-1]);
}
int queryi(int l,int r)
{
    int x=(int)(log((double)(r-l+1))/log(2.0));
    return minn(dpi[l][x],dpi[r-(1<<x)+1][x]);
}
int maxx(int x,int y)
{
    return a[x]>=a[y]?x:y;
}
void rmqa(int len)
{
    for(int i=0; i<len; i++)
    dpa[i][0]=i;
    for(int j=1; (1<<j)<len; j++)
    for(int i=0; i+(1<<j)-1<len; i++)
    dpa[i][j]=maxx(dpa[i][j-1],dpa[i+(1<<(j-1))][j-1]);
}
int querya(int l,int r)
{
    int x=(int)(log((double)(r-l+1))/log(2.0));
    return maxx(dpa[l][x],dpa[r-(1<<x)+1][x]);
}
int main()
{
    int x,y,q,i,t;
    while(~scanf("%d%d",&x,&q))
    {
        for(i=0;i<x;i++)
        scanf("%d",&a[i]);
        rmqi(x);
        rmqa(x);
        while(q--)
        {
            int l,r;
            scanf("%d%d",&l,&r);
            if(l>r)
            swap(l,r);
            printf("%d\n",a[querya(l-1,r-1)]-a[queryi(l-1,r-1)]);
        }
    }
}