POJ 3273 Monthly Expense（二分）

优化预算策略：最小化农场开支最大月度限额

原创于 2021-02-23 20:32:16 发布 · 212 阅读

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农夫John面临资金挑战，需将N天支出划分为M财政周期（fajomonths），目标是最大化最小月度支出。通过二分查找最小化最高月度花费。实例展示了解题步骤和代码实现。

Farmer John is an astounding accounting wizard and has realized he might run out of money to run the farm. He has already calculated and recorded the exact amount of money (1 ≤ moneyi ≤ 10,000) that he will need to spend each day over the next N (1 ≤ N ≤ 100,000) days.

FJ wants to create a budget for a sequential set of exactly M (1 ≤ M ≤ N) fiscal periods called “fajomonths”. Each of these fajomonths contains a set of 1 or more consecutive days. Every day is contained in exactly one fajomonth.

FJ’s goal is to arrange the fajomonths so as to minimize the expenses of the fajomonth with the highest spending and thus determine his monthly spending limit.

Input
Line 1: Two space-separated integers: N and M
Lines 2…N+1: Line i+1 contains the number of dollars Farmer John spends on the ith day
Output
Line 1: The smallest possible monthly limit Farmer John can afford to live with.
Sample Input

Sample Output

Hint
If Farmer John schedules the months so that the first two days are a month, the third and fourth are a month, and the last three are their own months, he spends at most $500 in any month. Any other method of scheduling gives a larger minimum monthly limit.

题意：
有n个数，划分为m个连续的区间，对于每一个区间内数的和，使其最大值尽可能小

思路：
最小化最大值，用二分解决。先找出下界L和上界R， while(L< R)，求出mid，看这个mid值是符合题意，继续二分。最后R即为答案。
对于每一个mid，遍历一遍n个数，看能划分为几个区间，如果划分的区间小于（或等于）m个，说明上界取大了，那么令R=mid，否则另 L=mid+1.

代码：

#include"stdio.h"
#include"string.h"
#include"algorithm"
using namespace std;
int a[100010];
int main()
{
	int m,n,i,L=-1,R=0;
	scanf("%d %d",&n,&m);
	for(i=1;i<=n;i++)
	{
		scanf("%d",&a[i]);
		if(L<a[i])
		L=a[i];//下界是n个数的最大值
		R+=a[i];//上界是n个数的和
	}
	while(L<R)
	{
		int mid=(L+R)/2,s=0,ans=0;
		for(i=1;i<=n;i++)
		{
			if(ans+a[i]>mid)
			{
				s++;
				ans=a[i];
			}
			else
			ans+=a[i];
		}
		s++;
		if(s<=m)
		R=mid;
		else
		L=mid+1;
	}
	printf("%d",R);
	return 0;
}