POJ3273-Monthly Expense

大致题意：

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

 

解题思路：

经典的二分穷举

详细的思路我写在程序注释中，这样会更容易懂

看完我的程序还是无法切入题目的同学，建议先用 朴素的穷举 去左这题，虽然很大机会会超时，但是只是为了辅助理解。本题的二分纯粹是一个优化穷举的工具。

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 nextN (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 andM
Lines 2.. N+1: Line i+1 contains the number of dollars Farmer John spends on theith day

Output

Line 1: The smallest possible monthly limit Farmer John can afford to live with.

Sample Input

7 5100400300100500101400

Sample Output

500

 1 //Memory Time  2 //612K   297MS  3  4 #include<iostream> 5 using namespace std; 6  7 int n; //天数 8 int m; //规定的分组数 9 10 /*判断用当前的mid值能把天数n分成几组*/11 /*通过比较group与m的大小，对mid值进行优化*/12 13 bool judge_group(int mid,int money[])14 {15     int sum=0;16     int group=1;    //当前mid值能把n天分成的组数(初始把全部天数作为1组)17 18     for(int i=1;i<=n;i++)  //从第一天开始向下遍历每天的花费19         if(sum+money[i]<=mid)  //当前i天之和<=mid时，把他们归并到一组20             sum+=money[i];21         else               //若 前i-1天之和 加上第i天的花费 大于mid22         {23             sum=money[i];  //则把前i-1天作为一组，第i天作为下一组的第一天24             group++;    //此时划分的组数+125         }26 27     if(group>m)28         return false;   //若利用mid值划分的组数比规定的组数要多，则说明mid值偏小29     else30         return true;    //否则mid值偏大31 }32 33 int main(void)34 {35     while(cin>>n>>m)36     {37         int* money=new int[n+1];  //每天花费的金额38         int low=0;  //下界39         int high=0; //上界40 41         for(int i=1;i<=n;i++)42         {43             cin>>money[i];44 45             high+=money[i];   //把所有天数的总花费作为上界high（相当于把n天都分作1组）46             if(low<money[i])47                 low=money[i]; //把n天中花费最多的那一天的花费作为下界low（相当于把n天分为n组）48         }                     //那么要求的值必然在[low,high]范围内49 50         int mid=(low+high)/2;51 52         while(low<high)  //可能在low==high之前，分组数就已经=m，但是mid并不是最优，因此要继续二分53         {54             if(!judge_group(mid,money))55                 low=mid+1;     //mid值偏小，下界前移56             else57                 high=mid-1;    //mid值偏大，上界后移58 59             mid=(low+high)/2;60         }61 62         cout<<mid<<endl;  //二分搜索最后得到的mid值必然是使得分组符合要求的最优值63 64         delete money;65     }66     return 0;67 }