zoukankan      html  css  js  c++  java
  • poj 3264 Balanced Lineup

                                                                                                                                                                   Balanced Lineup
    Time Limit: 5000MS   Memory Limit: 65536K
    Total Submissions: 51876   Accepted: 24319
    Case Time Limit: 2000MS

    Description

    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..N+Q+1: Two integers A and B (1 ≤ ABN), 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
    题意:给定一串长度为N的数列,现在给定子序列[a,b],要查询连续的子数列[a,b]区间中中的最大值和最小值的差。
    思路:典型的线段树问题,对于线段树中每个节点k,维护两个值,即维护该节点对应的区间[l,r)中的最大值和最小值,最后输出其差即可。
    AC代码:
    #define _CRT_SECURE_NO_DEPRECATE
    #include<iostream>
    #include<algorithm>
    #include<vector>
    using namespace std;
    const int ST_SIZE = (1 << 17) - 1,N_MAX=50000+2;
    int N, Q;
    int height[N_MAX];
    int dat_large[ST_SIZE], dat_small[ST_SIZE];
    void init(int k,int l,int r) {//节点k,对应区间[l,r)
        if (r - l == 1) {
           dat_large[k] = dat_small[k] = height[l];//!!!!!!!!!!
     }
        else {
            int left = 2 * k + 1;
            int right = 2 * k + 2;
            init(left,l,(l+r)/2);
            init(right, (l + r) / 2, r);
            dat_large[k] = max(dat_large[left],dat_large[right]);
            dat_small[k] = min(dat_small[left],dat_small[right]);
        }
    }
    
    pair<int,int> query(int k,int l,int r,int a,int b) {//节点k,对应区间[l,r),查找区间[a,b),用于找区间[a,b)的最大最小值
        //pair<int,int>find;//分别存放最大和最小值
        if (b <= l || a >= r) {//无交集
            return make_pair(0,INT_MAX);
        }
        else if (a <= l&& b>= r) {//完全包含区间!!!!!!!!!!!!!!!!!!
            return make_pair(dat_large[k], dat_small[k]);
         }
        else {
            pair<int, int>find1 = query(2*k+1,l,(l+r)/2,a,b);
            pair<int, int>find2 = query(2 * k + 2, (l + r) / 2, r, a, b);
            int large = max(find1.first,find2.first);
            int small = min(find1.second,find2.second);
            return make_pair(large, small);
        }
    }
    
    int main() {
        scanf("%d%d",&N,&Q);
        for (int i = 0; i < N;i++) {
            scanf("%d",&height[i]);
        }
        init(0,0,N);
        for (int i = 0; i < Q; i++) {
            int a, b;
            scanf("%d%d",&a,&b);
            a--, b--;
            pair<int, int>find = query(0,0,N,a,b+1);
            printf("%d
    ",find.first-find.second);
        }
        return 0;
    }
  • 相关阅读:
    Java实现监控目录下文件变化
    Postgresql 修改用户密码
    Swing清空jtable中的数据
    delphi登录用友的信息
    用友U8的SQL SERVER 数据库结构说明表
    候老师的讲堂:视频录制、笔记软件、思维导图、画图等工具
    DELPHI 关于内存数据与 JSON
    Delphi国内优秀网站及开源项目
    SQL Server 阻止了对组件Ad Hoc Distributed Queries访问的方法
    SQL Server跨服务器查询
  • 原文地址:https://www.cnblogs.com/ZefengYao/p/6690944.html
Copyright © 2011-2022 走看看