Hdu-5884 Sort (k Fork Huffman Tree)

Problem Description recently, Bob has just learnt a naive sorting algorithm:merge sort. Now, Bob receives a task from Alice.
Alice would 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 the 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 integers N (2≤n≤100000) and T (∑ni=1ai<t<231).
The next line there is 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
Source ACM/ICPC Asia Regional Qingdao Online
Analysis: Two +k fork Huffman Tree, note (n-1)% (k-1). = 0 O'Clock special treatment is needed.

#include <iostream> #include <string> #include <algorithm> #include <cstdlib> #include < cstdio> #include <set> #include <map> #include <vector> #include <cstring> #include <stack > #include <queue> #define INF 0x3f3f3f3f #define EPS 1e-9 #define MOD 1000000007 #define MAXN 200005 using Nam
Espace std;
typedef long Long LL;
int N,T,T,A[MAXN];
    void Nextint (int &x)//input plug-in {char C; Do C = GetChar ();  
    while (C < ' 0 ' | | c > ' 9 ');  
    x = C ' 0 ';  
while (' 0 ' <= (c = GetChar ()) && c<= ' 9 ') x = x*10 + C-' 0 ';
	} ll check (int mid) {deque <ll> q1,q2;
	for (int i = 1;i <= n;i++) q1.push_back (A[i]);
	int res = (n-1)% (mid-1);
	ll tmp = 0,ans = 0;
	if (res) res++;
		for (int i = 1;i <= res;i++) {tmp + = Q1.front ();  
	Q1.pop_front ();
	} ans + = tmp;
	if ((n-1)% (mid-1)) Q2.push_back (TMP);
		while (Q1.size () + q2.size () > 1) {ll tmp = 0; for (int i = 1;i <= mid;i++) {if (q2.empty () | |
				Q1.front () < Q2.front ()) && q1.size ()) {tmp + = Q1.front ();
			Q1.pop_front ();
				} else {tmp + = Q2.front ();
			Q2.pop_front ();
		}} ans + = tmp;
	Q2.push_back (TMP);
} return ans; 
	} int main () {nextint (T);
		while (t--) {nextint (n), Nextint (T);
		for (int i = 1;i <= n;i++) nextint (A[i]);
		Sort (a+1,a+1+n);
		int L = 2,r = n;
			while (l! = r) {int mid = (l+r)/2;
			if (check (mid) <= 1ll*t) R = Mid;	
		else L = mid+1;	
	} cout<<l<<endl; }
}