POJ 3273 Monthly Expense（最大化最小值）

最新推荐文章于 2020-02-02 13:44:33 发布

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最新推荐文章于 2020-02-02 13:44:33 发布

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本文介绍了一个预算划分问题，目标是最小化最高预算限制。通过将连续的n天分成m段连续区间来实现，每段的总花费不能超过设定的预算限制。采用二分查找算法寻找最优解，并提供了一个示例代码实现。

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题意：把连续的n天，分为连续的m段，使得每段中的花费尽可能的小，求最大的花费。

解题思路：典型的二分，可行性的判定模拟一下，因为是连续的所以a数组的次序不能打乱（不能排序）。

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.

Memory: 1088 KB		Time: 32 MS
Language: G++		Result: Accepted

#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<cctype>
#include<list>
#include<iostream>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>

using namespace std;

#define FOR(i, s, t) for(int i = (s) ; i <= (t) ; ++i)
#define REP(i, n) for(int i = 0 ; i < (n) ; ++i)

int buf[10];
inline long long read()
{
    long long x=0,f=1;
    char ch=getchar();
    while(ch<'0'||ch>'9')
    {
        if(ch=='-')f=-1;
        ch=getchar();
    }
    while(ch>='0'&&ch<='9')
    {
        x=x*10+ch-'0';
        ch=getchar();
    }
    return x*f;
}

inline void writenum(int i)
{
    int p = 0;
    if(i == 0) p++;
    else while(i)
        {
            buf[p++] = i % 10;
            i /= 10;
        }
    for(int j = p - 1 ; j >= 0 ; --j) putchar('0' + buf[j]);
}
/**************************************************************/
#define MAX_N 100005
const int INF = 0x3f3f3f3f;
int a[MAX_N];
int n, m;

bool f(int x)
{
    int cnt = 1;
    int sum = 0;
    for(int i = 0 ; i < n ; i++)
    {
        if(a[i] > x)
            return false;
        if(sum + a[i] <= x)
        {
            sum += a[i];
        }
        else
        {
            cnt++;
//            cout<<cnt<<endl;
//            cout<<sum<<endl;
            sum = a[i];
        }
        if(cnt > m) return false;
    }
    return true;
}


int main()
{
    while(~scanf("%d%d", &n, &m))
    {
        for(int i = 0 ; i < n ; i++)
        {
            a[i] = read();
        }
        int lb = -1, ub = 100005;
        while(ub - lb > 1)
        {
            int mid = (ub + lb) /2;
            if(f(mid)) ub = mid;
            else lb = mid;
        }
        printf("%d\n", ub);
    }

    return 0;
}