zoukankan      html  css  js  c++  java
  • Day6

    For the daily milking, Farmer John's N cows (1 ≤ N ≤ 50,000) always line up in the same order. One day Farmer John decides to organize a game of Ultimate Frisbee with some of the cows. To keep things simple, he will take a contiguous range of cows from the milking lineup to play the game. However, for all the cows to have fun they should not differ too much in height.

    Farmer John has made a list of Q (1 ≤ Q ≤ 200,000) potential groups of cows and their heights (1 ≤ height ≤ 1,000,000). For each group, he wants your help to determine the difference in height between the shortest and the tallest cow in the group.

    Input

    Line 1: Two space-separated integers, N and Q.
    Lines 2.. N+1: Line i+1 contains a single integer that is the height of cow i
    Lines N+2.. NQ+1: Two integers A and B (1 ≤ A ≤ B ≤ N), representing the range of cows from A to B inclusive.

    Output

    Lines 1.. Q: Each line contains a single integer that is a response to a reply and indicates the difference in height between the tallest and shortest cow in the range.

    Sample Input

    6 3
    1
    7
    3
    4
    2
    5
    1 5
    4 6
    2 2

    Sample Output

    6
    3
    0

    思路:ST表板子题,ST[i][j]表示下表从i到i+2^j-1的最值,查询时,已知l与r,长度len=r-l+1,且2^log2(len)>len/2,令k=log2(len),ST[l][k]肯定超过了长度的一半,反向取后侧,r-m+1=2^len,另一侧就是ST[r-2^k+1][k]
    const int maxm = 5e4+10;
    
    int Max[maxm][20], Min[maxm][20], N, Q;
    
    int main() {
        scanf("%d%d", &N, &Q);
        int t, l, r;
        for(int i = 1; i <= N; ++i) {
            scanf("%d", &t);
            Max[i][0] = Min[i][0] = t;
        }
        for(int k = 1; (1<<k) <= N; ++k) {
            for(int i = 1; i+(1<<k)-1 <= N; ++i) {
                Max[i][k] = max(Max[i][k-1], Max[i+(1<<(k-1))][k-1]);
                Min[i][k] = min(Min[i][k-1], Min[i+(1<<(k-1))][k-1]);
            }
        }
        for(int i = 0; i < Q; ++i) {
            scanf("%d%d", &l, &r);
            int k = log((double)(r-l+1)) / log(2.0);
            printf("%d
    ", max(Max[l][k],Max[r-(1<<k)+1][k]) - min(Min[l][k], Min[r-(1<<k)+1][k]));
        }
        return 0;
    }
    View Code
    
    
  • 相关阅读:
    jsp表单数据添加到数据库
    javaweb 复习随笔
    [组 原]
    [组 原]
    均方差、交叉熵及公式推导
    网络安全知识网站
    docker搭建渗透环境并进行渗透测试
    SQL注入之-DECLARE时间盲注
    Apache Flink Dashboard未授权访问导致任意Jar包上传漏洞
    Dnscat2实现DNS隐蔽隧道反弹Shell
  • 原文地址:https://www.cnblogs.com/GRedComeT/p/12198916.html
Copyright © 2011-2022 走看看