hdu 5933 ArcSoft's Office Rearrangement

最新推荐文章于 2019-09-22 10:40:26 发布

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本文介绍了一种针对ArcSoft公司员工分布的优化算法，旨在通过最少的操作步骤将不同规模的工作区块重新分配为相同规模的新区块。该算法考虑了合并与拆分两种操作，并通过实例演示了如何计算达到目标所需的最小操作数。

Problem Description
ArcSoft, Inc. is a leading global professional computer photography and computer vision technology company.

There are N working blocks in ArcSoft company, which form a straight line. The CEO of ArcSoft thinks that every block should have equal number of employees, so he wants to re-arrange the current blocks into K new blocks by the following two operations:

merge two neighbor blocks into a new block, and the new block’s size is the sum of two old blocks’.
split one block into two new blocks, and you can assign the size of each block, but the sum should be equal to the old block.

Now the CEO wants to know the minimum operations to re-arrange current blocks into K block with equal size, please help him.

Input
First line contains an integer T, which indicates the number of test cases.

Every test case begins with one line which two integers N and K, which is the number of old blocks and new blocks.

The second line contains N numbers a1, a2, ⋯, aN, indicating the size of current blocks.

Limits
1≤T≤100
1≤N≤105
1≤K≤105
1≤ai≤105

Output
For every test case, you should output ‘Case #x: y’, where x indicates the case number and counts from 1 and y is the minimum operations.

If the CEO can’t re-arrange K new blocks with equal size, y equals -1.

Sample Input

3
1 3
14
3 1
2 3 4
3 6
1 2 3

Sample Output

Case #1: -1
Case #2: 2
Case #3: 3

#include<iostream>
#include<cstdio>
using namespace std;
long long a[100002];
int main(){
    long long T,N,K,cas=0;
    scanf("%lld",&T);
    while(T--){
        scanf("%lld%lld",&N,&K);
        cas++;
        long long sum=0;
        for(int i=1;i<=N;i++){
            scanf("%lld",&a[i]);
            sum+=a[i];
        }
        if(sum%K!=0){
            printf("Case #%lld: -1\n",cas);
            continue;
        }
        long long avg=sum/K,k=0,op=0;
        for(int i=1;i<=N;i++){
            if(a[i]>avg){
                k=0;
                op+=a[i]/avg;
                k=a[i]%avg;
                if(k==0) op--;
                if(k!=0){
                    a[i+1]+=k;
                    op++;
                }
            }
            else if(a[i]==avg){
                continue;
            }
            else if(a[i]<avg){
                a[i+1]+=a[i];
                op++;
            }
        }
        printf("Case #%lld: %lld\n",cas,op);
    }
    return 0;
}