POJ 3273 - Monthly Expense（二分）

最新推荐文章于 2019-01-18 09:46:49 发布

原创最新推荐文章于 2019-01-18 09:46:49 发布 · 573 阅读

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本文介绍了一个经典的预算划分问题，即如何将连续的预算周期划分成若干个连续的时间段（fajomonths），使得每个时间段内的最大支出最小化。通过使用二分查找的方法，有效地解决了这个问题，并提供了完整的代码实现。

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 11178		Accepted: 4579

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.

=============================

题意：给出农夫每天的花费，连续分组使各组的花费之和尽量小，输出最大值

对最大值进行二分

#include <iostream>
#include <cstdio>

using namespace std;
int n,m,a[111111],MAX=-1,sum=0;

bool C(int x)
{
    int cnt=0;
    for(int i=-1;i<n;)
    {
        int ans=0;
        while(ans<=x)
        {
            i++;
            if(i==n) break;
            ans+=a[i];
            //cout<<a[i]<<" "<<i<<" "<<ans<<endl;
        }
        if(i!=n) i--;
        cnt++;
    }
    if(cnt<=m) return 1;
    else return 0;
}

int b_search()
{
    int l=MAX,r=sum,mid,ans=0;
    while(l<=r)
    {
        mid=(l+r)/2;
        if(C(mid))
        {
            ans=mid;
            r=mid-1;
        }
        else l=mid+1;
    }
    return ans;
}

int main()
{
    scanf("%d%d",&n,&m);
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        if(MAX<a[i]) MAX=a[i];
        sum+=a[i];
    }
    //cout<<MAX<<" "<<sum<<endl;
    int ans=b_search();
    printf("%d\n",ans);
    return 0;
}