POJ 3264 Balanced Lineup (rmq/segment tree)

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 40312		Accepted: 18936
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'll take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to has 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≤ H Eight ≤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 G Roup.

Input

Line 1:two space-separated integers, Nand Q.
Lines 2.. N+1:line I+1 contains a single integer which is the height of cow I
Lines N+2.. N+ Q+1:two integers Aand B(1≤ A≤ B≤ N), representing the range of cows from ATo BInclusive.

Output

Lines 1.. Q: Each line contains a single integer so is a response to a reply and indicates the difference in height between the Tal Lest and shortest cow in the range.

Sample Input

6 31734251 54) 62 2

Sample Output

630

Source

Test instructions: Give a non-ordered sequence of N, then m query, input two number l,r, output closed interval [l,r] maximum value and minimum worth difference.

Ideas

1. A segment tree can be used to do, a variable in the segment tree records the maximum value of an interval, the other variable records the minimum value of an interval, and then input l,r, respectively, the query, the result of the final subtraction is the required value.

2. You can also use RMQ to do, after the input sequence, preprocessing, and then each input l,r directly after the maximum and minimum value subtraction.

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Line segment Tree procedure:

#include <iostream> #include <algorithm> #include <stdio.h> #include <stdlib.h> #include <    string.h>using namespace std;struct node{int l;    int R;    int xx; int yy;}    Q[200010];int n,m;void Build (int l,int R,int rt) {q[rt].l = l;    Q[RT].R = R;        if (L = = r) {scanf ("%d", &q[rt].yy);        q[rt].xx = q[rt].yy;    return;    } int mid = (l+r) >>1;    Build (l,mid,rt<<1);    Build (mid+1,r,rt<<1|1);    q[rt].yy = min (q[rt<<1].yy,q[rt<<1|1].yy); q[rt].xx = max (q[rt<<1].xx,q[rt<<1|1].xx);}    int qurry1 (int ll,int rr,int l,int r,int RT) {if (ll<=l && rr>=r) {return q[rt].xx;    } if (Ll>r | | rr<l) {return 0;    } int mid = (l+r) >>1; Return Max (Qurry1 (ll,rr,l,mid,rt<<1), Qurry1 (ll,rr,mid+1,r,rt<<1|1));}    int qurry2 (int ll,int rr,int l,int r,int RT) {if (ll<=l && rr>=r) {return q[rt].yy;      } if (Ll>r | | rr<l) {  return 99999999;    } int mid = (l+r) >>1; return min (Qurry2 (ll,rr,l,mid,rt<<1), Qurry2 (ll,rr,mid+1,r,rt<<1|1));}        int main () {while (scanf ("%d%d", &n,&m)!=eof) {build (1,n,1);        int x, y;            for (int i=0;i<m;i++) {scanf ("%d%d", &x,&y);            int maxx = Qurry1 (x,y,1,n,1);            int minn = Qurry2 (x,y,1,n,1);        printf ("%d\n", Maxx-minn); }} return 0;}

RMQ Procedure:

#include <iostream> #include <algorithm> #include <stdio.h> #include <stdlib.h> #include < String.h> #include <math.h>using namespace std;const int N = 200010;int maxx[n][20];int minn[n][20];int n,m;void    Play () {int L = floor (log10 (double (n))/log10 (double (2))); for (int j=1;j<=l;j++) {for (int i=1;i<=n+1-(1&LT;&LT;J); i++) {Maxx[i][j] = max (maxx[i][j-1],maxx[i+ (1<< (j-1))]            [J-1]);        Minn[i][j] = min (minn[i][j-1],minn[i+ (1<< (j-1))][j-1]);        }}}int Main () {while (scanf ("%d%d", &n,&m)!=eof) {int x, y;            for (int i=1;i<=n;i++) {scanf ("%d", &x);            Maxx[i][0] = x;        Minn[i][0] = x;        } play ();            while (m--) {scanf ("%d%d", &x,&y);            int pp = floor (log10 (double (y-x+1))/log10 (double (2)));        printf ("%d\n", Max (maxx[x][pp],maxx[y-(1&LT;&LT;PP) +1][pp])-min (minn[x][pp],minn[y-(1&LT;&LT;PP) +1][pp]));   }    } return 0;} 

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POJ 3264 Balanced Lineup (rmq/segment tree)