POJ-3264 Balanced Lineup (RMQ problem seeking interval maximum)

RMQ (Range minimum/maximum Query) The question is: For series A of length n, answer a number of questions RMQ (A,I,J) (i,j<=n), and return to column A and subscript in I, The smallest (large) value in J, that is, the RMQ problem is the problem of finding the maximum interval.

Balanced Lineup

Time Limit: 5000MS

Memory Limit: 65536KB

64bit IO Format: %i64d &%i64u

Submit Status

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

1. Simplicity (traversal): Complexity O (n)-O (QN).

2. Segment Tree: Complexity O (n)-O (Qlogn).

3.ST (Sparse Table) algorithm: O (NLOGN)-O (q)

The ST algorithm, because each query only O (1), when processing a large number of queries there is an advantage.

<1>. preprocessing (Dynamic programming DP)

For A[i] series, F[i][j] indicates the maximum value (DP state) from the number of consecutive 2^j in the sequence of I, and you can see that f[i][0] represents A[i] (the initial value of DP). Finally, the state transition equation is

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

<2> Enquiry

If the query interval is (a, b), the interval length is b-a+1, and k=log2 (b-a+1) is taken, then Max (A, b) =max (F[a][k],f[b-2^k+1][k]).

1.ST algorithm

#include <iostream> #include <algorithm> #include <cstdio> #include <cstring> #include < Cmath> #define MAX (A, B) (a>b?a:b) #define MIN (A, b) (A&LT;B?A:B) using namespace std;const int maxn = 50050;int mins[m Axn][20];int maxs[maxn][20];void RMQ (int n) {for (int j = 1; (1 << j) <= n;j++) for (int i = 1; i + (1 << J)-1 <= N; i++) {int p = (1 << (j-1)); Mins[i][j] = Min (mins[i][j-1], mins[i + p][j-1]); Maxs[i][j] = max (maxs[i][j-1], maxs[i + p][j-1]);}} int querymin (int l, int r) {int k = log (double) (r-l + 1))/log (2.0); return min (Mins[l][k], Mins[r-(1 << k) + 1][k ]);} int Querymax (int l, int r) {int k = log (double) (r-l + 1))/log (2.0); return Max (Maxs[l][k], Maxs[r-(1 << k) + 1][k ]);}  int main () {int n, q;scanf ("%d%d", &n, &q), int num;for (int i = 1; I <= n; i++) {scanf ("%d", &num); Maxs[i][0] = mins[i][0] = num;} RMQ (n); int A, b;int ans;for (int i = 0; i < Q; i++) {scanf ("%d%d", &a, &b); ans= Querymax (A, b)-Querymin (A, B);p rintf ("%d\n", ans);} 

2. Segment Tree

#include <iostream> #include <cstdio> #include <cstring> #include <string> #include < Algorithm> #define MAX (A, B) (a>b?a:b) #define MIN (A, b) (A&LT;B?A:B) using namespace std;const int maxn = 50050;int num [Maxn];struct node{int r;int l;int max;int Min;} Tree[3*maxn];void Build (int l, int r, int i) {tree[i].l = l; tree[i].r = r;if (L = = r) {Tree[i]. Max = Tree[i]. Min = Num[l];return;} int m = (L + r) >> 1, ls = i << 1, rs = ls + 1;build (l, M, LS); build (M + 1, R, RS); Tree[i]. max = max (Tree[rs]. Max, Tree[ls]. MAX); Tree[i]. min = min (Tree[rs]. Min, Tree[ls]. Min);} int Querymax (int l, int r, int i) {if (tree[i].l = = L&AMP;&AMP;TREE[I].R = = r) return Tree[i]. Max;int m = (tree[i].l + tree[i].r) >> 1, ls = i << 1, rs = ls + 1;if (r <= m) return Querymax (L, R, LS); El Se if (L > m) return Querymax (L, R, RS), else return Max (Querymax (L, M, ls), Querymax (M + 1, R, RS));} int querymin (int l, int r, int i) {if (tree[i].l = = L&AMP;&AMP;TREE[I].R = = r) return Tree[i]. MIn;int m = (tree[i].l + tree[i].r) >> 1, ls = i << 1, rs = ls + 1;if (r <= m) return Querymin (L, R, LS); Els E if (L > m) return querymin (L, R, RS), else return min (Querymin (L, M, ls), querymin (M + 1, R, RS));} int main () {int n, q;scanf ("%d%d", &n, &q), for (int i = 1; I <= n; i++) scanf ("%d", &num[i]); build (1, N, 1); int A, b;int ans;for (int i = 0; i < Q; i++) {scanf ("%d%d", &a, &b); ans = Querymax (A, B, 1)-Querymin (A, B, 1); printf ("%d\n", ans);}}

Reference to the http://blog.csdn.net/niushuai666/article/details/6624672/

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POJ-3264 Balanced Lineup (RMQ problem seeking interval maximum)