zoukankan      html  css  js  c++  java
  • hdu5443 The Water Problem

    The Water Problem

    Time Limit: 1500/1000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
    Total Submission(s): 602    Accepted Submission(s): 485


    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,...,106}. 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
     
    Source
     
    Recommend
    hujie   |   We have carefully selected several similar problems for you:  5449 5448 5447 5446 5445
    区间求最大值
    次裸裸的RMQ,套上模板即过
    #include<stdio.h>
    #include<string.h>
    #include<iostream>
    #include<math.h>
    #include<algorithm>
    using namespace std;
    #define N 50005
    int dpmin[N][20],dpmax[N][20];
    int main()
    {    
        int t;
        scanf("%d",&t);
        while(t--){
        int i, j, n, m;
        scanf("%d",&n);
        memset(dpmin,0,sizeof(dpmin));
        memset(dpmax,0,sizeof(dpmax));
        for(i=1; i<=n; i++)
        {
            scanf("%d", &dpmin[i][0]);
            dpmax[i][0]=dpmin[i][0];
        }
        int mm=floor(log(1.0*n)/log(2.0));
        for(j=1; j<=mm; j++)
            for(i=1; i<=n; i++)
            {
                if((i+(1<<(j-1)))<=n)
                {
        //            dpmin[i][j]=min(dpmin[i][j-1], dpmin[i+(1<<(j-1))][j-1]);
                    dpmax[i][j]=max(dpmax[i][j-1], dpmax[i+(1<<(j-1))][j-1]);
                }
            }
        int x, y;
            
            scanf("%d",&m);
        for(int i=1; i<=m; i++)
        {
            scanf("%d%d", &x, &y);
            int  mid=floor(log(y*1.0-x+1)/log(2.0));
            int maxnum=max(dpmax[x][mid], dpmax[y-(1<<mid)+1][mid]);
        //    int minnum=min(dpmin[x][mid], dpmin[y-(1<<mid)+1][mid]);
            printf("%d
    ", maxnum);
        }
        }
        return 0;
    }
  • 相关阅读:
    Extjs4 关于设置form中所有子控件为readOnly属性的解决方案
    Chrome调试(转)
    ExtJS4 动态加载
    CSS display和visibility的用法和区别
    利用Java调用OpenCV进行人脸识别
    Mac上安装openCV(Java版本)
    关于mysql的Fetch Time 和 Duration Time
    数据库相关中间件介绍
    JVM和java应用服务器调优
    探索 ConcurrentHashMap 高并发性的实现机制
  • 原文地址:https://www.cnblogs.com/13224ACMer/p/4812142.html
Copyright © 2011-2022 走看看