Monthly Expense POJ - 327（二分）

最新推荐文章于 2020-05-03 21:18:49 发布

我不过是个神秘人

最新推荐文章于 2020-05-03 21:18:49 发布

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本文链接：https://blog.youkuaiyun.com/weixin_43326028/article/details/94605748

本文介绍了一种使用二分查找算法来优化农场主约翰的预算规划策略的方法。面对连续N天的开支预测，约翰的目标是将这些日子划分为M个连续的财政周期，每个周期称为“fajomonth”，以确定每月的最低支出限制。通过二分查找每个月的最小可能花费，并检查该预算是否可行，最终确定了每月的最低预算限额。

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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
7 5
100
400
300
100
500
101
400
Sample Output
500
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.

二分每个月最小的花费，判断这个花费符合不符合条件。

#include<iostream>
#include<algorithm>
#include<stdio.h>
#include<set>
#include<map>
#include<queue>
#include<vector>
#include<string.h>
#include<cmath>
#define LL long long
#define mem(a, b) memset(a, b, sizeof(a))
#define N 100001
#define MOD
using namespace std;
int n, m, day[N];

bool judge(int x) {
	int t=0, tt=1;
	for(int i=1; i<=n; i++) { 
		if(day[i]>x) return 0;
		if(t+day[i]<=x) {
			t+=day[i];
		}
		else{
			t=day[i], tt++; 
		}
		if(tt>m) return 0;
	}
	return 1;
}

int main() {
	while(scanf("%d%d",&n, &m)!=EOF) {
		int l=0x3f3f3f3f, r=0;
		for(int i=1; i<=n; i++) {
			scanf("%d",&day[i]);
			l=min(l, day[i]);
			r+=day[i];
		}
		while(l<r) {
			int mid= (l+r)>>1;
			if(judge(mid)) {
				r=mid;
			}
			else l=mid+1;
		}
		cout<<l<<endl;
	}
}