Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 42929		Accepted: 20184
Case Time Limit: 2000MS

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.

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.

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.

6 3

1

7

3

4

2

5

1 5

4 6

2 2

6

3

0

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

using namespace std;

#define N 50010

int a[N];

int max_dp[N][];

int min_dp[N][];

int MAX(int i,int j){

    if(i>=j) return i;

    return j;

}

int MIN(int i,int j){

    if(i<=j) return i;

    return j;

}

void init_MAX_RMQ(int n){

    for(int i=;i<=n;i++) max_dp[i][]=a[i];

    for(int j=;(<<j)<=n;j++){

        for(int i=;i<=n-(<<j)+;i++){

            ///F[i, j]=max（F[i，j-1], F[i + 2^(j-1)，j-1]）。

            max_dp[i][j] = MAX(max_dp[i][j-],max_dp[i+(<<(j-))][j-]);

        }

    }

}

int MAX_RMQ(int a,int b){

    int k = (int)(log(b-a+1.0)/log(2.0));

    ///RMQ(A, i, j)=min{F[i,k],F[j-2^k+1,k]}

    return MAX(max_dp[a][k],max_dp[b-(<<k)+][k]);

}

void init_MIN_RMQ(int n){

    for(int i=;i<=n;i++) min_dp[i][]=a[i];

    for(int j=;(<<j)<=n;j++){

        for(int i=;i<=n-(<<j)+;i++){

            min_dp[i][j] = MIN(min_dp[i][j-],min_dp[i+(<<(j-))][j-]);

        }

    }

}

int MIN_RMQ(int a,int b){

    int k = (int)(log(b-a+1.0)/log(2.0));

    return MIN(min_dp[a][k],min_dp[b-(<<k)+][k]);

}

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m)!=EOF){

        for(int i=;i<=n;i++){

            scanf("%d",&a[i]);

        }

        init_MAX_RMQ(n);

        init_MIN_RMQ(n);

        while(m--){

            int a,b;

            scanf("%d%d",&a,&b);

            printf("%d\n",MAX_RMQ(a,b)-MIN_RMQ(a,b));

        }

    }

    return ;

}

#include <iostream>

#include <cstdio>

#include <cstring>

#include <cmath>

using namespace std;

#define N 50010

struct Tree{

    int l,r;

    int Max,Min;

}tree[*N];

int a[N];

int MAX_VALUE;

int MIN_VALUE;

int MAX(int i,int j){

    if(i>=j) return i;

    return j;

}

int MIN(int i,int j){

    if(i<=j) return i;

    return j;

}

void PushUp(int idx){

    tree[idx].Max = MAX(tree[idx<<].Max,tree[idx<<|].Max);

    tree[idx].Min = MIN(tree[idx<<].Min,tree[idx<<|].Min);

}

void build(int l,int r,int idx){

    tree[idx].l = l;

    tree[idx].r = r;

    if(l==r) {

        tree[idx].Max = tree[idx].Min = a[l];

        return ;

    }

    int mid=(l+r)>>;

    build(l,mid,idx<<);

    build(mid+,r,idx<<|);

    PushUp(idx);

}

void query(int l,int r,int idx){

    if(tree[idx].l==l&&tree[idx].r==r){

        MAX_VALUE = MAX(MAX_VALUE,tree[idx].Max);

        MIN_VALUE = MIN(MIN_VALUE,tree[idx].Min);

        return;

    }

    int mid=(tree[idx].l+tree[idx].r)>>;

    if(mid>=r) query(l,r,idx<<);

    else if(mid<l) query(l,r,idx<<|);

    else{

        query(l,mid,idx<<);

        query(mid+,r,idx<<|);

    }

}

int main()

{

    int n,m;

    while(scanf("%d%d",&n,&m)!=EOF){

        for(int i=;i<=n;i++){

            scanf("%d",&a[i]);

        }

        build(,n,);

        while(m--){

            int b,c;

            scanf("%d%d",&b,&c);

            MAX_VALUE=-;

            MIN_VALUE=;

            query(b,c,);

            printf("%d\n",MAX_VALUE-MIN_VALUE);

        }

    }

    return ;

}