ACM最大值最小化&&二分法

Description

 

Before the invention of book-printing, it was very hard to make a copy of a book. All the contents had to be re-written by hand by so calledscribers. The scriber had been given a book and after several months he finished its copy. One of the most famous scribers lived in the 15th century and his name was Xaverius Endricus Remius Ontius Xendrianus (Xerox). Anyway, the work was very annoying and boring. And the only way to speed it up was to hire more scribers.

 

Once upon a time, there was a theater ensemble that wanted to play famous Antique Tragedies. The scripts of these plays were divided into many books and actors needed more copies of them, of course. So they hired many scribers to make copies of these books. Imagine you have m books (numbered ) that may have different number of pages ( ) and you want to make one copy of each of them. Your task is to divide these books among k scribes, . Each book can be assigned to a single scriber only, and every scriber must get a continuous sequence of books. That means, there exists an increasing succession of numbers  such that i-th scriber gets a sequence of books with numbers between b_i-1+1 and b_i. The time needed to make a copy of all the books is determined by the scriber who was assigned the most work. Therefore, our goal is to minimize the maximum number of pages assigned to a single scriber. Your task is to find the optimal assignment.

 

Input 

The input consists of N cases. The first line of the input contains only positive integer N. Then follow the cases. Each case consists of exactly two lines. At the first line, there are two integers m and k, . At the second line, there are integers  separated by spaces. All these values are positive and less than 10000000.

 

Output 

For each case, print exactly one line. The line must contain the input succession  divided into exactly k parts such that the maximum sum of a single part should be as small as possible. Use the slash character (`/') to separate the parts. There must be exactly one space character between any two successive numbers and between the number and the slash.

 

If there is more than one solution, print the one that minimizes the work assigned to the first scriber, then to the second scriber etc. But each scriber must be assigned at least one book.

 

Sample Input 

2
9 3
100 200 300 400 500 600 700 800 900
5 4
100 100 100 100 100

 

Sample Output 

100 200 300 400 500 / 600 700 / 800 900
100 / 100 / 100 / 100 100

 

 解题思路：

题目大意是给定m个数字要求将他们划分成k个部分，得到每个区间最大值的最小值。我们可以用分治和二分的思想来完成这道题目。具体的可以看代码，上面有解释。

#include<cstdio>
#include<cstring>
long  T,m,k,a[510],b[510],ans[510],left,right,mid;
int  check(long max)
{
    long  j,num=0,tot=0;
    memset(b,0,sizeof(b));
    for(j=m;j>=1;j--)//由后往前查找'/'可摆放处
    {
        if(tot+a[j]<=max&&j>=k-num)//判断a[j]处能否继续摆放'/',,需要满足相加只和小于max和还剩元素个数大于等于我们还要划分的次数
            tot+=a[j];
        else
        {
            ++num;//切割次数记录
            b[j]=1;//测试设定'/'摆放标志
			tot=a[j];
        }
    }
    if(num+1==k)//判断测试结果是否满足条件，分成k份我们需要划k-1次
    {
        for(j=1;j<=m;j++)
            ans[j]=b[j];//标记
        return 1;
    }
    else return 0;
}
int main()
{
   long i;
    scanf("%lld",&T);//案例数
    while(T--)
    {
        left=0;
        right=0;
        scanf("%lld%lld",&m,&k);
        for(i=1;i<=m;i++)
        {
            scanf("%lld",&a[i]);
            right+=a[i];//将序列的和设为子序列和的最大限值
            if(a[i]>left) 
				left=a[i];//将序列中最大的值设为子序列的最小限值
        }
        memset(ans,0,sizeof(ans));//清0
        while(left<=right)//二分法查找最小的子序列的和
        {
            mid=(left+right)/2;
            if(check(mid)==1)
                right=mid-1;
            else left=mid+1;
        }
        for(i=1;i<m;i++)//输出控制
        {
            if(ans[i]==1)
                printf("%lld / ",a[i]);
            else printf("%lld ",a[i]);
        }
        printf("%lld\n",a[m]);
    }
    return 0;
}

　　

程序代码：