LightOJ - 1082    区间最小值

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 to J.

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).

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.

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

Dataset is huge. Use faster I/O methods.

这题是查询最小值  可以用线段数，也可以用RMQ算法，因为没有更新的操作，所以用RMQ的预处理，和访问即可

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<cmath>
using namespace std;
int n;
int dp[100010][50];
void rmq()
{
    for(int j=1; (1<<j)<=n; j++)
    {
        for(int i=1; i+(1<<j)-1<=n; i++)
        {
            dp[i][j]=min(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
        }
    }
}
int rmq1(int l,int r)
{


    int k=(int) log2((double)(r-l+1));
    return min(dp[l][k],dp[r-(1<<k)+1][k]);
}


int main()
{
    int t,k=1;
    scanf("%d",&t);
    while(t--)
    {
        printf("Case %d:\n",k++);
        scanf("%d",&n);
        int q;
        scanf("%d",&q);
        for(int j=1; j<=n; j++)
        {
            scanf("%d",&dp[j][0]);
        }
        rmq();
        while(q--)
        {
            int a,b;
            scanf("%d%d",&a,&b);
            printf("%d\n",rmq1(a,b));
        }
    }
    return 0;
}

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