lightoj-1082 - Array Queries

Posted 2020-07-16 FireCool

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 to J.

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

 

思路: 可用线段树，  但用分块来求，容易写，也不容易出现错误。

 

#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
using namespace std;

const int N = 100100,inf = 1000000;
int a[N];
int size[N/2];
int main(){
    
    int T,n,q,under1,under2,m,uu1,uu2,minn;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        memset(size,10,sizeof(size));
        scanf("%d%d",&n,&q);
        m = sqrt(n);
        for(int i=1;i<=n;i++){
            scanf("%d",&a[i]);
            size[(int)i/m] = min(size[(int)i/m],a[i]);
        }
        
        printf("Case %d:\\n",t);
        for(int i=0;i<q;i++){
            scanf("%d%d",&under1,&under2);
            uu1 = (int)under1/m;
            uu2 = (int)under2/m;
            minn = inf;
            for(int j = under1;j<=min((uu1+1)*m-1,under2);j++){
                minn = min(minn,a[j]);
            }
            for(int j = uu1+1;j<uu2;j++){
                minn = min(minn,size[j]);
            }
            if(uu1!=uu2){
                for(int j=uu2*m;j<=under2;j++) minn = min(minn,a[j]);
            }
            printf("%d\\n",minn);
        }
        
        
    }
    return 0;
}

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