hdoj 5443 The Water Problem 【RMQ】

The Water Problem Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others) Total Submission(s): 343    Accepted Submission(s): 281 Problem Description In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with  a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing  2  integers  l  and  r , please find out the biggest water source between  al  and  ar .   Input First you are given an integer  T(T≤10)  indicating the number of test cases. For each test case, there is a number  n(0≤n≤1000)  on a line representing the number of water sources.  n  integers follow, respectively  a1,a2,a3,...,an , and each integer is in  {1,...,106} . On the next line, there is a number  q(0≤q≤1000)  representing the number of queries. After that, there will be  q  lines with two integers  l  and  r(1≤l≤r≤n)  indicating the range of which you should find out the biggest water source.   Output For each query, output an integer representing the size of the biggest water source.   Sample Input 3 1 100 1 1 1 5 1 2 3 4 5 5 1 2 1 3 2 4 3 4 3 5 3 1 999999 1 4 1 1 1 2 2 3 3 3   Sample Output 100 2 3 4 4 5 1 999999 999999 1

RMQ裸题

ac代码：

#include <cstdio>
#include <cstring>
#include <cmath>
#include <algorithm>
#define MAXN 10010
#define INF 1000000000
using namespace std;
int A[MAXN];
int N, M;
int Amax[MAXN][50];
void RMQ_init()
{
    for(int i = 1; i <= N; i++)
        Amax[i][0] = A[i];
    for(int j = 1; (1<<j) <= N; j++)
    {
        for(int i = 1; i + (1<<j)-1 <= N; i++)
            Amax[i][j] = max(Amax[i][j-1], Amax[i+(1<<(j-1))][j-1]);
    }
}
int query(int L, int R)
{
    int k = 0;
    while((1<<(k+1)) <= R-L+1) k++;
    return max(Amax[L][k], Amax[R-(1<<k)+1][k]);
}
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d", &N);
        for(int i = 1; i <= N; i++)
            scanf("%d", &A[i]);
        RMQ_init();
        scanf("%d", &M);
        int x, y;
        while(M--)
        {
            scanf("%d%d", &x, &y);
            printf("%d\n", query(x, y));
        }
    }
    return 0;
}