RMQ问题 acm.mipt 105题解

题目：

RMQ problem
Time limit = 5 second(s)

Memory limit = 32000 Kb

You are given large array of real numbers:
a[0], a[1], ..., a[N-1]

Range Minimal Query problem:

Problem RMQ(i,j) = "find minimum of a[i], a[i+1], ..., a[j-1]".

Your program should solve given set of RMQ problems.

Input. Input data has the following format:

N
a[0] a[1] ... a[N-1]
M
i1 j1
i2 j2
...

iM jM

Here N ≤ 250000, M ≤ 500000, 0 ≤ ik < N, 0 < jk ≤ N, ik < jk.

Output. Your program should output M numbers b1, b2, ..., bM delimited by space, where

bk = MIN (a[ik], a[ik + 1], ..., a[jk — 1]).

Input#1
10
1 2 3 4 5 6 7 8 9 10
16
0 1
0 2
0 3
0 4
3 4
3 5
3 6
3 7
3 8
3 9
3 10
0 10
9 10
8 10
7 10
5 6
Output#1
1.000000
1.000000
1.000000
1.000000
4.000000
4.000000
4.000000
4.000000
4.000000
4.000000
4.000000
1.000000
10.000000
9.000000
8.000000
6.000000

Input#2
10
3.86934 7.28362 2.15556 14.75963 0.33240 17.12550 -0.71121 13.90834 -1.13470 5.99831
11
6 10
0 1
5 10
1 9
0 6
0 10
2 5
3 10
5 9
0 8
2 10
Output#2
-1.134700
3.869340
-1.134700
-1.134700
0.332400
-1.134700
0.332400
-1.134700
-1.134700
-0.711210
-1.134700

代码：
[cpp]
#include <stdio.h>
#include <stdlib.h>
#include <math.h>
using namespace std;

#define N 250010
int n,m;
int x,y;
float A[N];
int M[N][19];

void process()
{
    for(int i=0;i<n;i++)
    {
        M[i][0] = i;
    }
    for(int j=1;(1<<j)<=n;j++)
    {
        for(int i=0;i+(1<<j)-1<n;i++)
        {
            if(A[M[i][j-1]]<A[M[i+(1<<(j-1))][j-1]])
            {
                M[i][j] = M[i][j-1];
            }
            else
            {
                M[i][j] = M[i+(1<<(j-1))][j-1];
            }
        }
    }
}

float query(int x,int y)
{
    int k;
    k = (int)(log(y-x+1)/log(2));

    return A[M[x][k]]<A[M[y-(1<<k)+1][k]] ? A[M[x][k]] : A[M[y-(1<<k)+1][k]];
}

int main()
{
    /*#ifndef ONLINE_JUDGE
        freopen("in.txt","r",stdin);
    #endif*/
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        scanf(" %f",&A[i]);
    }
    process();
    scanf(" %d",&m);

    for(int i=0;i<m;i++)
    {
        scanf(" %d%d",&x,&y);
        printf("%f\n",query(x,y-1));
    }
    return 0;

}

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