POJ 3273 Monthly Expense

最新推荐文章于 2020-02-02 13:44:33 发布

原创最新推荐文章于 2020-02-02 13:44:33 发布 · 407 阅读

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本文介绍了一个预算规划问题，通过二分查找算法确定最低的月度支出限制。农民约翰需要规划连续的财政周期，每个周期包含连续的天数，并且每天只属于一个周期。目标是最小化最高支出周期的成本。

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题目链接：http://poj.org/problem?id=3273

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21599		Accepted: 8486

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

思路：因为答案最小要为它们中的最大值，最大要为它们的和，所以可以二分答案，尽量分给前面未分够的，然后看次数是否满足条件，不断缩小区间，求出最小值。直接上代码吧。

附上AC代码：

#include <cstdio>
//#pragma comment(linker, "/STACK:102400000, 102400000")
using namespace std;
const int maxn = 100005;
int cost[maxn];
int n, m;

int main(){
	#ifdef LOCAL
	freopen("input.txt", "r", stdin);
	freopen("output.txt", "w", stdout);
	#endif
	while (~scanf("%d%d", &n, &m)){
        int left=0, mid=0, right=0;
        for (int i=0; i<n; ++i){
            scanf("%d", cost+i);
            right += cost[i];
            if (cost[i] > left)
                left = cost[i];
        }
        int cnt=0, sum=0;
        while (right > left){
            mid = (right+left)>>1;
            cnt=0, sum=0;
            for (int i=0; i<n; ++i){
                if (sum+cost[i] <= mid)
                    sum += cost[i];
                else
                    ++cnt, sum=cost[i];
            }
            if (cnt+1 > m)
                left = mid+1;
            else
                right = mid-1;
        }
        printf("%d\n", left);
	}
	return 0;
}