poj 3264 RMQ

Posted 2020-07-12 半根毛线code

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 44075		Accepted: 20687
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

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

Sample Output

6
3
0

Source

USACO 2007 January Silver

 

题意： 求区间最大最小值 输出差值

 

题解： RMQ问题 裸题   dp思想+位运算

     

          基础问题码一遍  f[i][j] 代表 以第i个数为起点 长度为2^j次方的区间的最小值

          查询操作中的以左右边界为起点取并集 输出ans

 1 #include<iostream>
 2 #include<cstring>
 3 #include<cstdio>
 4 #include<queue>
 5 #include<stack>
 6 #include<vector>
 7 #include<map>
 8 #define ll __int64
 9 using namespace std;
10 int n,q;
11 int f[50005][30];
12 int ff[50005][30];
13 int l,r;
14 void rmq_init()
15 {
16     for(int j=1;(1<<j)<=n;j++)
17     for(int i=1;i+(1<<(j))-1<=n;i++)
18     {
19         f[i][j]=min(f[i][j-1],f[i+(1<<(j-1))][j-1]);
20         ff[i][j]=max(ff[i][j-1],ff[i+(1<<(j-1))][j-1]);
21         //cout<<f[i][j]<<endl;
22     }
23     
24 }
25 int rmq(int aa,int bb)
26 {
27     int k=0;
28     int ans1,ans2;
29     while((1<<(k+1))<=bb-aa+1)
30     k++;
31     ans1=min(f[aa][k],f[bb-(1<<k)+1][k]);
32     ans2=max(ff[aa][k],ff[bb-(1<<k)+1][k]);
33     return ans2-ans1;
34 }
35 int main()
36 {
37     while(scanf("%d %d",&n,&q)!=EOF)
38     {
39         for(int i=1;i<=n;i++)
40             {
41             scanf("%d",&f[i][0]);
42             ff[i][0]=f[i][0];
43             }
44         rmq_init();
45         for(int i=1;i<=q;i++)
46          {
47            scanf("%d %d",&l,&r);
48            cout<<rmq(l,r)<<endl;
49          }    
50     }
51     return 0;
52  }