POJ3273-Monthly Expense

最新推荐文章于 2019-04-20 09:48:00 发布

Wang_128

最新推荐文章于 2019-04-20 09:48:00 发布

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分类专栏： POJ ----二分

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本文介绍了一个经典的预算划分问题，即如何将连续的每日开销划分成若干个时间段（fajomonths），使得开销最高的时间段的金额最小。通过使用二分查找的方法，有效地解决了这一问题，并提供了完整的代码实现。

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Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 25981		Accepted: 10027

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.

Source

USACO 2007 March Silver

题意：共n个月，给出每个月的开销，将n个月划分成m个时间段，求m个时间段中开销最大的时间段的最小开销值

解题思路：二分

#include <iostream>
#include <cstdio>
#include <cstring>
#include <string>
#include <algorithm>
#include <cmath>
#include <map>
#include <cmath>
#include <set>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <functional>

using namespace std;

#define LL long long
const int INF=0x3f3f3f3f;

int a[100009];
int n,m;

bool check(int k)
{
    int sum=0,cnt=0;
    for(int i=1;i<=n;i++)
    {
        if(a[i]>k) return 0;
        if(sum+a[i]<k) sum+=a[i];
        else if(sum+a[i]==k) {sum=0,cnt++;}
        else {sum=a[i];cnt++;}
    }
    if(sum) cnt++;
    if(cnt<=m) return 1;
    else return 0;
}

int main()
{
	while(~scanf("%d%d",&n,&m))
    {
        for(int i=1;i<=n;i++) scanf("%d",&a[i]);
        int l=0,r=INF,ans;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(check(mid)) {r=mid-1,ans=mid;}
            else l=mid+1;
        }
        printf("%d\n",ans);
    }
	return 0;
}