Balanced Lineup RMQ 中的ST算法(第一次做)

Problem 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 ≤ ABN), 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 *********************************************************************************************************************************** dp[i][j]表示在区间(i,i+2^j-1)中所取得最值; 其中求最值时也有优点dp,每次把最值保存在dp数组中 ST算法最主要的优点是查询的时候费时为o(1); ***********************************************************************************************************************************
 1 #includeiostream
 2 #includestring
 3 #includecstring
 4 #includecmath
 5 #includequeue
 6 #includestack
 7 #includemap
 8 #includecstdio
 9 using namespace std;
10 int cow[51001];
11 int n,m,p;
12 int dpmax[51001][51],dpmin[51001][51];
13 void maxmp(int n)
14 {
15     for(int it=1; it=n; it++)
16         dpmax[it][0]=cow[it];
17     for(int it=1; it=(log((double)n)/log(2.0)); it++)
18         for(int jt=1; jt+(1it)-1=n; jt++)
19             dpmax[jt][it]=max(dpmax[jt][it-1],dpmax[jt+(1(it-1))][it-1]);
20 }
21 
22 void minmp(int n)
23 {
24     for(int it=1; it=n; it++)
25         dpmin[it][0]=cow[it];
26     for(int it=1; it=(log((double)n)/log(2.0)); it++)
27         for(int jt=1; jt+(1it)-1=n; jt++)
28             dpmin[jt][it]=min(dpmin[jt][it-1],dpmin[jt+(1(it-1))][it-1]);
29 }
30 
31 int querymax(int s,int v)
32 {
33     int k=(int)((log((double)(v-s+1))/log(2.0)));
34     return max(dpmax[s][k],dpmax[v-(1k)+1][k]);
35 }
36 
37 int querymin(int s,int v)
38 {
39     int k=(int)((log((double)(v-s+1))/log(2.0)));
40     return min(dpmin[s][k],dpmin[v-(1k)+1][k]);
41 }
42 
43 int main()
44 {
45 
46     while(~scanf("%d %d",n,p))
47     {
48         for(int i = 1 ; i = n ; ++i)
49             scanf("%d",cow[i]);
50 
51         maxmp(n);
52         minmp(n);
53         while(p--)
54         {
55             int s,v;
56             scanf("%d%d",s,v);
57             printf("%d\n",querymax(s,v)-querymin(s,v));
58         }
59     }
60     return 0;
61 
62 }
View Code

 

转载于:https://www.cnblogs.com/sdau--codeants/p/3483738.html

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