hdu 5884 Sort 2016ACM/ICPC青岛赛区网络赛1007

Problem Description

Recently, Bob has just learnt a naive sorting algorithm: merge sort. Now, Bob receives a task from Alice.

Alice will give Bob N sorted
sequences, and the i-th
sequence includes ai elements.
Bob need to merge all of these sequences. He can write a program, which can merge no more than k sequences
in one time. The cost of a merging operation is the sum of the length of these sequences. Unfortunately, Alice allows this program to use no more than T cost.
So Bob wants to know the smallest k to
make the program complete in time.

 

Input

The first line of input contains an integer t0,
the number of test cases. t0 test
cases follow.

For each test case, the first line consists two integers N (2≤N≤100000) and T (∑Ni=1ai<T<231).

In the next line there are N integers a1,a2,a3,...,aN(∀i,0≤ai≤1000).

 

Output

For each test cases, output the smallest k.

 

Sample Input

1
5 25
1 2 3 4 5

 

Sample Output

3

二分k

然后就是k叉的哈夫曼树

我们可以直接把原数列sort一下

然后用两个数列维护。每次计算结果放第二个数列中，再每次比较两格数列的头指针的数哪个小，累加

因为数组可以在二分外面直接排序，这样就比直接用堆做少了一个log

#include<map>
#include<cmath>
#include<queue>
#include<vector>
#include<cstdio>
#include<string>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
struct number
{
long long x;
bool operator <(number y) const
{
return x>y.x;
}
}b[100001];
int a[100001];
int n;
long long t;
priority_queue<number> Q;
long long q1[1000001],q2[1000001];
inline bool check(int limit)
{
int i,j;
int l1=1,r1=0,l2=1,r2=0;
int m=limit-1;
int tt=n;
while(tt%m!=1&&m!=1)
{
r1++;
q1[r1]=0;
tt++;
}
for(i=1;i<=n;i++)
{
r1++;
q1[r1]=a[i];
}
long long sum=0,ans=0;
/* for(i=1;i<=((n-1)-1)%m+1;i++)
{ vv
l1++;
sum+=q1[l1];
}*/
ans+=sum;
// r2++;
// q2[r2]=sum;
for(i=1;i<=tt/m;i++)
{
sum=0;
int sx=0;
while(sx<limit&&l1<=r1&&l2<=r2)
{
if(q1[l1]<=q2[l2])
{
sum+=q1[l1];
l1++;
}
else
{
sum+=q2[l2];
l2++;
}
sx++;
}
if(sx<limit)
{
if(l1<=r1)
{
while(sx<limit)
{
sum+=q1[l1];
l1++;
sx++;
}
}
else
{
while(sx<limit)
{
sum+=q2[l2];
l2++;
sx++;
}
}
}
r2++;
q2[r2]=sum;
ans+=sum;
}
return ans<=t;
}
int main()
{
int T;
scanf("%d",&T);
while(T>0)
{
T--;
scanf("%d%I64d",&n,&t);
int i;
for(i=1;i<=n;i++)
scanf("%d",&a[i]);
sort(a+1,a+1+n);
for(i=1;i<=n;i++)
b[i].x=a[i];
int l=2,r=n;
while(l<=r)
{
int mid=(l+r)/2;
if(check(mid))
r=mid-1;
else
l=mid+1;
}
printf("%d\n",l);
}
return 0;
}