poj 3273 Monthly Expense (二分)

最新推荐文章于 2017-08-11 11:21:18 发布

原创最新推荐文章于 2017-08-11 11:21:18 发布 · 562 阅读

0 ·

CC 4.0 BY-SA版权

计算方法专栏收录该内容

2 篇文章

订阅专栏

本文介绍了一个经典的预算划分问题：如何将连续的每日开销划分为多个连续区间（称为“fajomonth”），使得这些区间的最大总开销最小。通过二分查找与贪心策略结合的方法，有效地解决了这个问题。

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 19990		Accepted: 7884

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

给出农夫在n天中每天的花费，要求把这n天分作m组，每组的天数必然是连续的，要求分得各组的花费之和应该尽可能地小，最后输出各组花费之和中的最大值

#include <iostream>
#include <cstring>
#include <cstdio>
#include <algorithm>
#include <queue>
#include <stack>
#define LL long long 
#define INF 0x3f3f3f3f
#define eps 1e-8

using namespace std;

int n,m;

bool Judge(int mid, int *arr)
{
	int ant = 1;
	int sum = 0;
   	for(int i=0;i<n;i++)
	{
		if(arr[i]+sum <= mid)
		{
			sum += arr[i];
		}
		else
		{
			sum = arr[i];
			ant++;
		}
	}
	if(ant <= m)
		return true;
	return false;

}

int main()
{
	while(~scanf("%d%d",&n,&m))
	{
		int *arr = new int[n+1];
		int Low = 0;
		int High = 0;
	   	for(int i=0;i<n;i++)
		{
			scanf("%d",&arr[i]);
			High += arr[i];
			Low = max(Low, arr[i]);
		}
		int mid = (Low + High)>>1;	
		while(Low < High)
		{
		//	int mid  exactly = (Low + High)>>1;
			if(!Judge(mid,arr))
				Low = mid + 1;
			else
				High = mid - 1;
			mid = (Low + High) >> 1;
		}
		printf("%d\n",mid);
	}
	return 0;
}