Saturday, December 5, 2015

最大值最小化问题 - Bisection - binary search


http://blog.csdn.net/nanjunxiao/article/details/8145971
把一个包含n个正整数的序列划分成m个连续的子序列。设第i个序列的各数之和为S(i),求所有S(i)的最大值最小是多少?
例如序列1 2 3 2 5 4划分为3个子序列的最优方案为 1 2 3 | 2 5 | 4,其中S(1),S(2),S(3)分别为6,7,4,那么最大值为7;
如果划分为 1 2 | 3 2 | 5 4,则最大值为9,不是最小。

问题分析:
能否使m个连续子序列所有的s(i)均不超过x,则该命题成立的最小的x即为答案。该命题不难判断,只需贪心,每次尽量从左
向右尽量多划分元素即可。
我们把该问题转化为递归分治问题,类似于二分查找。首先取Sum和元素最大值的中值x,如果命题为假,那么答案比x大;
如果命题为真,则答案小于等于x。问题得解,复杂度为O(n*logSum)

bool is_part(int x)//是否能把序列划分为每个序列之和不大于x的m个子序列 
{
 //每次往右划分,划分完后,所用的划分线不大于m-1个即可
 int t=0,s=0;
 bool ok=true;
 for(int i=0;i<n;i++)
 {
  if(A[i]>x)//假如有其中一个元素大于x,则极端地把划分线分别设在在其左边和右边,都不能使这一个只有一个元素的序列之和不大于x,更不用说所有序列之和不大于x 
  {
   ok=false;
   break;
  }
  if(s+A[i]>x)//大于,不能再把当前元素加上了 
  {
   t++;//多用了一条横杠
   s=A[i];
   if(t>m-1)//t=m时退出,即在最后一个元素之前都已经用了m条划分线,此时已把序列划分成了m+1个序列,太过分了,让其适可而止 
   {
    ok=false;break;
   }
  }else
  {
   s+=A[i];//把当前元素与前面的元素连上,以便尽量往右划分,贪心到底 
  }
 }
 return ok;
}
int sum()
{
 int s=0;
 for(int i=0;i<n;i++) s+=A[i];
 return s;
}
int binary_solve()
{
 int x=max,y=sum(); 
 while(x<y)
 {
  int m=x+(y-x)/2;
  if(is_part(m)) y=m;
  else x=m+1;
 }
 return x;
}
http://blog.csdn.net/lzw_java/article/details/7486512


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