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  • ACM学习历程—HDU 5443 The Water Problem(RMQ)(2015长春网赛1007题)

    Problem Description
    In Land waterless, water is a very limited resource. People always fight for the biggest source of water. Given a sequence of water sources with a1,a2,a3,...,an representing the size of the water source. Given a set of queries each containing 2 integers l and r , please find out the biggest water source between al and ar .
     
    Input
    First you are given an integer T(T10) indicating the number of test cases. For each test case, there is a number n(0n1000) on a line representing the number of water sources. n integers follow, respectively a1,a2,a3,...,an , and each integer is in {1,...,10^6} . On the next line, there is a number q(0q1000) representing the number of queries. After that, there will be q lines with two integers l and r(1lrn) indicating the range of which you should find out the biggest water source.
     
    Output
    For each query, output an integer representing the size of the biggest water source.
     
    Sample Input
    3
    1
    100
    1
    1 1
    5
    1 2 3 4 5
    5
    1 2
    1 3
    2 4
    3 4
    3 5
    3
    1 999999 1
    4
    1 1
    1 2
    2 3
    3 3
     
    Sample Output
    100
    2
    3
    4
    4
    5
    1
    999999
    999999
    1

    题目大意就是给定区间,求区间最值。

    这里采用了RMQ的ST算法,直接套的模板。

    代码:

    #include <iostream>
    #include <cstdio>
    #include <cstdlib>
    #include <cstring>
    
    using namespace std;
    
    const int maxN = 1005;
    int n, q;
    int a[maxN];
    int ma[maxN][20];
    
    void RMQ()
    {
        memset(ma, 0, sizeof(ma));
        for (int i = 0; i < n; ++i)
            ma[i][0] = a[i];
        for (int j = 1; (1<<j) <= n; ++j)
            for (int i = 0; i+(1<<j)-1 < n; ++i)
                ma[i][j] = max(ma[i][j-1], ma[i+(1<<(j-1))][j-1]);
    }
    
    int query(int lt, int rt)
    {
        int k = 0;
        while ((1<<(k+1)) <= rt-lt+1)
            k++;
        return max(ma[lt][k], ma[rt-(1<<k)+1][k]);
    }
    
    void input()
    {
        scanf("%d", &n);
        for (int i = 0; i < n; ++i)
            scanf("%d", &a[i]);
        RMQ();
    }
    
    void work()
    {
        scanf("%d", &q);
        int u, v, ans;
        for (int i = 0; i < q; ++i)
        {
            scanf("%d%d", &u, &v);
            ans = query(u-1, v-1);
            printf("%d
    ", ans);
        }
    }
    
    int main()
    {
        //freopen("test.in", "r", stdin);
        int T;
        scanf("%d", &T);
        for (int times = 0; times < T; ++times)
        {
            input();
            work();
        }
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/andyqsmart/p/4808216.html
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