zoukankan      html  css  js  c++  java
  • HDU 5289 Assignment rmq

    Assignment

    题目连接:

    http://acm.hdu.edu.cn/showproblem.php?pid=5289

    Description

    Tom owns a company and he is the boss. There are n staffs which are numbered from 1 to n in this company, and every staff has a ability. Now, Tom is going to assign a special task to some staffs who were in the same group. In a group, the difference of the ability of any two staff is less than k, and their numbers are continuous. Tom want to know the number of groups like this.

    Input

    In the first line a number T indicates the number of test cases. Then for each case the first line contain 2 numbers n, k (1<=n<=100000, 0<k<=10^9),indicate the company has n persons, k means the maximum difference between abilities of staff in a group is less than k. The second line contains n integers:a[1],a[2],…,an,indicate the i-th staff’s ability.

    Output

    For each test,output the number of groups.

    Sample Input

    2
    4 2
    3 1 2 4
    10 5
    0 3 4 5 2 1 6 7 8 9

    Sample Output

    5
    28

    Hint

    题意

    给你n个数,然后连续的,且这个区间内最大值减去最小值的差小于k的话,就可以算作一队。

    问你一共有多少种分队的方法。

    题解:

    暴力枚举左端点位置,然后二分枚举右端点,用一个rmq维护一下就好了。

    代码

    #include<bits/stdc++.h>
    using namespace std;
    const int maxn = 1e5+7;
    struct RMQ{
        const static int RMQ_size = maxn;
    	int n;
    	int ArrayMax[RMQ_size][21];
    	int ArrayMin[RMQ_size][21];
    
    	void build_rmq(){
    		for(int j = 1 ; (1<<j) <= n ; ++ j)
    			for(int i = 0 ; i + (1<<j) - 1 < n ; ++ i){
    				ArrayMax[i][j]=max(ArrayMax[i][j-1],ArrayMax[i+(1<<(j-1))][j-1]);
    				ArrayMin[i][j]=min(ArrayMin[i][j-1],ArrayMin[i+(1<<(j-1))][j-1]);
    			}
    	}
    
    	int QueryMax(int L,int R){
    		int k = 0;
    		while( (1<<(k+1)) <= R-L+1) k ++ ;
    		return max(ArrayMax[L][k],ArrayMax[R-(1<<k)+1][k]);
    	}
    
    	int QueryMin(int L,int R){
    		int k = 0;
    		while( (1<<(k+1)) <= R-L+1) k ++ ;
    		return min(ArrayMin[L][k],ArrayMin[R-(1<<k)+1][k]);
    	}
    
    
    	void init(int * a,int sz){
    		n = sz ;
    		for(int i = 0 ; i < n ; ++ i) ArrayMax[i][0] = ArrayMin[i][0] = a[i];
    		build_rmq();
    	}
    
    }solver;
    int a[maxn];
    int main()
    {
        int t;
        scanf("%d",&t);
        while(t--)
        {
            int n,k;scanf("%d%d",&n,&k);
            for(int i=0;i<n;i++)
                scanf("%d",&a[i]);
            solver.init(a,n);
            long long ans = 0;
            for(int i=0;i<n;i++)
            {
                int l = i,r = n-1,Ans=i;
                while(l<=r)
                {
                    int mid = (l+r)/2;
                    if(solver.QueryMax(i,mid)-solver.QueryMin(i,mid)<k)l=mid+1,Ans=mid;
                    else r=mid-1;
                }
                ans += (Ans-i+1);
            }
            cout<<ans<<endl;
        }
    	return 0;
    }
  • 相关阅读:
    如何在ONENET云端搭建IOT平台
    01_接口测试介绍
    10_fiddler_待整理
    09_fiddler_慢网络测试(限制网速)
    08_Fiddler_打断点(bpu)
    07_Fiddler_post提交到主体的四种参数形式
    06_Fiddler_get请求(url详解)
    05_Fiddler的Script 脚本用法
    04_Fiddler_Composer创建和发送HTTP Request
    03_Fiddler抓包的捕获设置
  • 原文地址:https://www.cnblogs.com/qscqesze/p/5269577.html
Copyright © 2011-2022 走看看