Balanced Lineup+POJ+RMQ问题

Balanced Lineup

Time Limit: 5000MS		Memory Limit: 65536K
Total Submissions: 34739		Accepted: 16352
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

解决方案：相当于RMQ的模板题，O(nlog(n))预处理后，可实现O(1)查询。

代码如下：
#include<iostream>
#include<cstdio>
#include<cmath>
#define MMAX 50005
using namespace std;
int maxdp[MMAX][30];
int mindp[MMAX][30];
int A[MMAX];
void initRMQ(int len)
{
    for(int i=1; i<=len; i++) mindp[i][0]=maxdp[i][0]=A[i];
    for(int j=1; (1<<j)<=len; j++)
        for(int i=1; i+(1<<j)-1<=len; i++)
        {
            maxdp[i][j]=max(maxdp[i][j-1],maxdp[i+(1<<(j-1))][j-1]);
            mindp[i][j]=min(mindp[i][j-1],mindp[i+(1<<(j-1))][j-1]);
        }

}
int RMQ(int L,int R)
{
    int k=0;
    while((1<<(k+1))<=(R-L+1))k++;
    int res=max(maxdp[L][k],maxdp[R-(1<<k)+1][k])-min(mindp[L][k],mindp[R-(1<<k)+1][k]);
    return res;
}
int main()
{
    int n,q;
    while(~scanf("%d%d",&n,&q))
    {
        for(int i=1; i<=n; i++)
        {
            scanf("%d",&A[i]);
        }
        initRMQ(n);
        for(int i=0; i<q; i++)
        {
            int L,R;
            scanf("%d%d",&L,&R);
            printf("%d\n",RMQ(L,R));
        }
    }
    return 0;
}