POJ 3273 Monthly Expense

最新推荐文章于 2023-09-01 22:00:12 发布

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最新推荐文章于 2023-09-01 22:00:12 发布

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分类专栏： ACM_二分文章标签：二分查找 poj

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本文介绍了一个经典的时间复杂度为O(n log n)的预算划分问题，通过二分枚举的方法来解决如何将连续的几天划分为若干组以使得每组的最大开销最小的问题。

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Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 15876		Accepted: 6346

Description

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 ≤ money_i ≤ 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.

题目大意：已知n天的花费金额，将n天分为m组，连续的几天才能组成一组，使各组的花费尽量小，求m个组中的最大花费金额。

解题思路：经典的二分枚举，以几组中最大花费金额mid为变量，合并后花费金额小于mid的天数组合，组数为判断条件，求出答案。

解题过程：wa点：写判断条件时要注意，因为花费要尽量小，所以mid值要尽量小。

代码如下：

#include <cstdio>
const int maxn=100010;
int a[maxn];
int main()
{
    int n,m,i,mid,high=0,low=0,sum,t;
    scanf("%d%d",&n,&m);
    for(i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        high+=a[i];
        if(low<a[i])
            low=a[i];
    }
    mid=(high+low)/2;
    while(low<high)
    {
		t=1,sum=0;
        for(i=0;i<n;i++)
        {
            if(sum+a[i]<=mid)
                sum+=a[i];
			else
			{
				sum=a[i];
				t++;
			}
        }
        if(t>m) 
            low=mid+1;
		else
            high=mid-1;
        mid=(high+low)/2;
    }
    printf("%d\n", mid);
	return 0;
}