POJ 3264 Balanced Lineup 【ST表静态RMQ】-优快云博客

传送门：http://poj.org/problem?id=3264

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 66241		Accepted: 30833
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 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

Source

USACO 2007 January Silver

题意概括：

给出一段长度为 N 的序列，和 Q 次查询。

每次输入区间，求解该区间最大值 - 最小值的结果。

解题思路：

RMQ问题可以线段树维护（甚至树状数组）复杂度预处理 O（NlongN）单次查询 O（logN）

不过这里用的是 ST表预处理O（NlogN）单次查询 O（1）

ST表本质思想是 dp，这里用两个dp 维护区间最大值和最小值。

设起点是 i 区间长度为 2 ^j

区间【i ， i +（1 << j )】的最大值为 dpmax[ i, j ]，最小值为 dpmin[ i, j ]；

转移方程：

dpmax[ i, j ] = max( dpmax[ i ][ j-1 ], dpmax[ i + (1<<(j-1)) ][ j ] );

dpmin[ i, j ] = min( dpmin[ i ][ j-1 ], dpmin[ i + (1<<(j-1)) ][ j ] );

实质就是按照二的幂次方关系，把一个区间分成了两个区间。

AC code：

 1 #include <cstdio>
 2 #include <iostream>
 3 #include <cstring>
 4 #include <cmath>
 5 #define INF 0x3f3f3f3f
 6 #define LL long long
 7 using namespace std;
 8 const int MAXN = 5e4+50;
 9 int dpmax[MAXN][20];
10 int dpmin[MAXN][20];
11 int num[MAXN];
12 
13 void make_maxRMQ(int N, int b[])
14 {
15     for(int i = 0; i < N; i++){
16         dpmax[i][0] = b[i];
17     }
18     for(int ilen = 1; (1<<ilen) <= N; ilen++)
19     for(int i = 0; i+(1<<ilen)-1 < N; i++){
20         dpmax[i][ilen] = max(dpmax[i][ilen-1], dpmax[i+(1<<(ilen-1))][ilen-1]);
21     }
22 }
23 
24 int get_max(int ll, int rr)
25 {
26     int k = (int)(log(rr-ll+1.0)/log(2.0));
27     return max(dpmax[ll][k], dpmax[rr-(1<<k)+1][k]);
28 }
29 
30 void make_minRMQ(int N, int a[])
31 {
32     for(int i = 0; i < N; i++){
33         dpmin[i][0] = a[i];
34     }
35     for(int ilen = 1; (1<<ilen) <= N; ilen++)
36     for(int i = 0; i+(1<<ilen)-1 < N; i++){
37         dpmin[i][ilen] = min(dpmin[i][ilen-1], dpmin[i+(1<<(ilen-1))][ilen-1]);
38     }
39 }
40 
41 int get_min(int ll, int rr)
42 {
43     int k = (int)(log(rr-ll+1.0)/log(2.0));
44     return min(dpmin[ll][k], dpmin[rr-(1<<k)+1][k]);
45 }
46 
47 int main()
48 {
49     int N, Q;
50     int L, R;
51     int ans;
52     while(~scanf("%d%d", &N, &Q)){
53         for(int i = 0; i < N; i++){
54             scanf("%d", &num[i]);
55         }
56         make_maxRMQ(N, num);
57         make_minRMQ(N, num);
58 
59         while(Q--){
60             scanf("%d%d", &L, &R);
61             L--, R--;
62             ans = get_max(L, R) - get_min(L, R);
63             printf("%d\n", ans);
64         }
65     }
66     return 0;
67 }