Allocation Aizu - ALDS1_4_D
You are given n packages of wi kg from a belt conveyor in order (i=0,1,…n−1). You should load all packages onto k trucks which have the common maximum load P. Each truck can load consecutive packages (more than or equals to zero) from the belt conveyor unless the total weights of the packages in the sequence does not exceed the maximum load P.
Write a program which reads n, k and wi, and reports the minimum value of the maximum load P to load all packages from the belt conveyor.
Input
In the first line, two integers n and k are given separated by a space character. In the following n lines, wi are given respectively.
Output
Print the minimum value of P in a line.
Constraints
1≤n≤100,000
1≤k≤100,000
1≤wi≤10,000
Sample Input 1
5 3
8
1
7
3
9
Sample Output 1
10
If the first truck loads two packages of {8,1}, the second truck loads two packages of {7,3} and the third truck loads a package of {9}, then the minimum value of the maximum load P shall be 10.
Sample Input 2
4 2
1
2
2
6
Sample Output 2
6
If the first truck loads three packages of {1,2,2} and the second truck loads a package of {6}, then the minimum value of the maximum load P shall be 6.
思路
确定最大运载量P时，首先要编写一个算法来计算k辆以内的卡车总共能装多少货物。
思路很简单，只要卡车的运载量没达到P，就让其继续按顺序装货物，最后再计算所有卡车运载量的总和即可。
以P为实参，编写一个返回可装载货物数v的函数v=f§，这个函数的算法复杂度为O(n)。
现在只需要调用这个函数，让P从0开始逐渐自增，第一个让v大于等于n的P就是答案。但是逐个检查P会失败的原因算法复杂度达到O(Pn)，程序不可能在限制时间内完成处理。
利用“P增加v不会减少”的性质，用二分搜索来求P，算法复杂度降低至O(n*log P)。
code
/*
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*/
// Virtual_Judge —— Allocation Aizu - ALDS1_4_D.cpp created by VB_KoKing on 2019-05-04:11.
/* Procedural objectives：
Variables required by the program:
Procedural thinking：
Functions required by the program:
*/
/* My dear Max said：
"I like you,
So the first bunch of sunshine I saw in the morning is you,
The first gentle breeze that passed through my ear is you,
The first star I see is also you.
The world I see is all your shadow."
FIGHTING FOR OUR FUTURE！！！
*/
#include <iostream>
#define MAX 100007
using namespace std;
int n,k;
long long T[MAX];
//k辆最大运载量为p的卡车能装多少货物
int check(long long p)
{
int i=0;
for (int j = 0; j < k; j++) {
long long s=0;
while (s+T[i]<p+1)
{
s+=T[i];
i++;
if (i==n)
return n;
}
}
return i;
}
int solve()
{
long long left=0;
long long right=100000*10000; //货物数*1件货物的最大重量
long long mid;
while (right-left>1)
{
mid=(right+left)/2;
int v=check(mid);
if (v>n-1) right=mid;
else left=mid;
}
return right;
}
int main()
{
cin>>n>>k;
for (int i = 0; i < n; i++)
cin>>T[i];
long long ans=solve();
cout<<ans<<endl;
return 0;
}
