POJ3273 Monthly Expense 二分

原创于 2019-01-24 10:21:45 发布 · 140 阅读

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本文探讨了如何通过二分搜索算法解决预算划分问题，旨在最小化最大预算支出，适用于农场主约翰的资金管理场景。文章提供了详细的算法实现，并附有样例输入输出，帮助读者理解并实践这一优化策略。

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 nextN (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

7 5100400300100500101400

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：这题看了好久都没看懂题意，就是将序列划分为k段，要求找出k段中最大的资金和要在所有符合要求的划分中值最少

思路：我们可以直到这个值必然在所有数中最大值与所有数的总和之间，那么只要再这个区间进行二分即可

#include <stdio.h>#include <string.h>#include <algorithm>using namespace std;int a[100005];int main(){    int n,m;    int sum,maxn,i,j,s,cnt,mid;    while(~scanf("%d%d",&n,&m))    {        sum = 0,maxn = 0;        for(i = 0; i<n; i++)        {            scanf("%d",&a[i]);            sum+=a[i];            maxn = max(maxn,a[i]);        }        while(maxn<sum)        {            mid = (sum+maxn)>>1;            s = 0,cnt = 0;            for(i = 0; i<n; i++)            {                s+=a[i];                if(s>mid)                {                    s = a[i];                    cnt++;                }            }            if(cnt<m) sum = mid;            else maxn = mid+1;        }        printf("%d\n",maxn);    }    return 0;}