[ACM] POJ 3273 Monthly Expense （二分解决最小化最大值）

Monthly Expense

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 14158		Accepted: 5697

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

7 5
100
400
300
100
500
101
400

Sample Output

500

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

解题思路：

题意为给定一个n个数组成的序列，划分为m个连续的区间，每一个区间全部元素相加，得到m个和，m个和里面肯定有一个最大值，我们要求这个最大值尽可能的小。

用二分查找能够非常好的解决问题。这类问题的框架为，找出下界left和上界right， while(left< right)， 求出mid，看这个mid值是符合题意，继续二分。最后right即为答案。

本题中的下界为n个数中的最大值，由于这时候，是要划分为n个区间（即一个数一个区间），left是满足题意的n个区间和的最大值，上届为全部区间的和，由于这时候，是要划分为1个区间（全部的数都在一个区间里面）,    1<=m<=n， 所以我们所要求的值肯定在 [left, right] 之间。对于每个mid，遍历一遍n个数，看能划分为几个区间，假设划分的区间小于（或等于）给定的m，说明上界取大了， 那么 另 right=mid，否则另 left=mid+1.

代码：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
const int maxn=100010;
int money[maxn];
int n,m;

int main()
{
    scanf("%d%d",&n,&m);
    int left=-1,right=0;
    for(int i=1;i<=n;i++)
    {
        scanf("%d",&money[i]);
        if(left<money[i])
            left=money[i];
        right+=money[i];
    }
    while(left<right)
    {
        int mid=(left+right)/2;
        int cnt=0;
        int cost=0;
        for(int i=1;i<=n;i++)
        {
            if(cost+money[i]>mid)
            {
                cnt++;//划分区间，不包含当前的money[i]
                cost=money[i];
            }
            else
                cost+=money[i];
        }
        cnt++;//最后一个cost值也要占一天
        if(cnt<=m)
            right=mid;
        else
            left=mid+1;
    }
    cout<<right<<endl;
    return 0;
}