RMQ优化解HDU3486

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>

using namespace std;
const int maxn=2e5+10;
int dp[maxn][18];
int n;

void st(int n)
{
    for(int j=1;j<=18;j++)
        {
            for(int i=1;i+(1<<(j-1))-1<=n;i++)
                {
                    dp[i][j]=max(dp[i][j-1],dp[i+(1<<(j-1))][j-1]);
                }
        }
}

int rmq(int i,int j)
{
    int k=0;
    while(1<<(k+1)<=j-i+1) k++;//位运算应该比log快，但博主还没有数据
    int ans=max(dp[i][k],dp[j-(1<<k)+1][k]);
    return ans;
}

int main()
{
    int k;
    while(~scanf("%d%d",&n,&k))
        {
            int sum=0,cnt=n,Max=-1;
            if(n<0&&k<0)
                break;
            for(int i=1;i<=n;i++)
                {
                    scanf("%d",&dp[i][0]);
                    if(dp[i][0]>Max) Max=dp[i][0];//求最大值
                    sum+=dp[i][0];
                }
            if(sum<=k)
                {
                    printf("-1
");
                    continue;
                }
            st(n);
            int beg=k/Max;//至少需要的组数，这步非常重要
            if(beg==0) beg=1;
            bool jg=false;
            for(int i=beg;i<n;i++)
                {
                    sum=0;
                    int t=n/i;段长度
                    for(int j=1;j<=i;j++)
                        {
                            sum+=rmq((j-1)*t+1,j*t);
                            if(sum>k)   break;
                        }
                    if(sum>k)
                        {
                            cnt=i;
                            break;
                        }
                }
                printf("%d
",cnt);
        }
    return 0;
}
--------------------- 
作者：BlueDoorZz 
来源：CSDN 
原文：https://blog.csdn.net/BlueDoorZz/article/details/81265144 
版权声明：本文为博主原创文章，转载请附上博文链接！

Problem Description

YaoYao has a company and he wants to employ m people recently. Since his company is so famous, there are n people coming for the interview. However, YaoYao is so busy that he has no time to interview them by himself. So he decides to select exact m interviewers for this task.
YaoYao decides to make the interview as follows. First he queues the interviewees according to their coming order. Then he cuts the queue into m segments. The length of each segment is , which means he ignores the rest interviewees (poor guys because they comes late). Then, each segment is assigned to an interviewer and the interviewer chooses the best one from them as the employee.
YaoYao’s idea seems to be wonderful, but he meets another problem. He values the ability of the ith arrived interviewee as a number from 0 to 1000. Of course, the better one is, the higher ability value one has. He wants his employees good enough, so the sum of the ability values of his employees must exceed his target k (exceed means strictly large than). On the other hand, he wants to employ as less people as possible because of the high salary nowadays. Could you help him to find the smallest m?

Input

The input consists of multiple cases.
In the first line of each case, there are two numbers n and k, indicating the number of the original people and the sum of the ability values of employees YaoYao wants to hire (n≤200000, k≤1000000000). In the second line, there are n numbers v1, v2, …, vn (each number is between 0 and 1000), indicating the ability value of each arrived interviewee respectively.
The input ends up with two negative numbers, which should not be processed as a case.

Output

For each test case, print only one number indicating the smallest m you can find. If you can’t find any, output -1 instead.

Sample Input

11 300 7 100 7 101 100 100 9 100 100 110 110 -1 -1

Sample Output

3

Hint

We need 3 interviewers to help YaoYao. The first one interviews people from 1 to 3, the second interviews people from 4 to 6, and the third interviews people from 7 to 9. And the people left will be ignored. And the total value you can get is 100+101+100=301>300.

题目大意：

有N名面试者，现在想通过分段录取最少的人，满足各人的能力值之和大于K.

解法：

运用RMQ算法，枚举，查询各区间最大值再求和，直至满足题目要求.题目思路简单，但很容易超时，必须对其进行优化.