Light oj 1082 - Array Queries 【RMQ 裸题】

1082 - Array Queries

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Time Limit: 3 second(s)Memory Limit: 64 MB

Given an array with N elements, indexed from 1 to N. Now you will be given some queries in the form I J, your task is to find the minimum value from index I toJ.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

The first line of a case is a blank line. The next line contains two integers N (1 ≤ N ≤ 10⁵), q (1 ≤ q ≤ 50000). The next line contains N space separated integers forming the array. There integers range in [0, 10⁵].

The next q lines will contain a query which is in the form I J (1 ≤ I ≤ J ≤ N).

Output

For each test case, print the case number in a single line. Then for each query you have to print a line containing the minimum value between index I and J.

Sample Input

Output for Sample Input

2

 

5 3

78 1 22 12 3

1 2

3 5

4 4

 

1 1

10

1 1

Case 1:

1

3

12

Case 2:

10

Note

Dataset is huge. Use faster I/O methods.

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PROBLEM SETTER: JANE ALAM JAN

题意：给你一个N个数组成的序列和Q次查询。每次查询区间[x,,y]最小值。

RMQ裸题，因为犯二RE了一次。。。

AC代码：

#include <cstdio>#include <cstring>#include <cmath>#include <cstdlib>#include <queue>#include <stack>#include <vector>#include <map>#include <string>#include <algorithm>#define MAXN 100000+10#define MAXM 2000+10#define INF 0x3f3f3f3fusing namespace std;int Amin[MAXN][30];int A[MAXN];int N, Q;void RMQ_init(){    for(int i = 1; i <= N; i++)        Amin[i][0] = A[i];    for(int j = 1; (1<<j) <= N; j++)    {        for(int i = 1; i + (1<<j)-1 <= N; i++)            Amin[i][j] = min(Amin[i][j-1], Amin[i+(1<<(j-1))][j-1]);    }}int query(int L, int R){    int k = 0;    while((1<<(k+1)) <= R-L+1) k++;    return min(Amin[L][k], Amin[R-(1<<k)+1][k]);}int k = 1;int main(){    int t;    scanf("%d", &t);    while(t--)    {        scanf("%d%d", &N, &Q);        for(int i = 1; i <= N; i++)            scanf("%d", &A[i]);        RMQ_init();        int a, b;        printf("Case %d:\n", k++);        while(Q--)        {            scanf("%d%d", &a, &b);            printf("%d\n", query(a, b));        }    }    return 0;}