HDU 5884 Sort (Huffman tree + two minutes)

In fact, this is Huffman tree, a start with priority queue write time-out, and later found that no priority queue, because each time the added value is always greater than before, then only need to use the normal queue can be, and this queue is monotonically increment, and then take the original array of the head or the head of the monotonous queue can be.

But without taking into account (n-1)% (k-1) ==0, the feeling that this condition is more difficult to think about, need to implement a preprocessing merge once.

Feeling can be derived by mathematical induction.

Main.cpp//Richard////Created by Shaojinjie on 16/9/22. COPYRIGHT©2016 year Shaojinjie.
All rights reserved. #include <iostream> #include <cstdio> #include <cstring> #include <algorithm> #include <
Queue> using namespace std;
typedef long Long LL;
const int maxn=200000+100;
int VAL[MAXN];
int n;
ll T;
    bool Merge (int k) {ll ans=0;
    int head=1,tail=n;
    Queue<ll> Q;
        if (k!=1&& (n-1)% (k-1)!=0) {for (int i=1;i<= (n-1)% (k-1) +1;i++) ans+=val[i];
        head= (n-1)% (k-1) +2;
    Q.push (ANS); } while (head<=tail| |!
        Q.empty ()) {int sum=0; for (int i=0;i<k;i++) {if (Head<=tail&&!q.empty () &&val[head]<q.front ()) {sum+=
            val[head++];}
            else if (!q.empty () &&head<=tail&&q.front () <=val[head]) {Sum+=q.front (); Q.pop ();}
            else if (Q.empty ()) {sum+=val[head++];} else if (head>tail) {sum+=q.front (); Q.Pop ();}
        } ans+=sum;
        if (ans>t) break;
        Q.push (sum);
    if (Q.size () ==1&&head>tail) break;
} return ans>t?false:true;
    } int main () {int t;
    scanf ("%d", &t);
        while (t--) {scanf ("%d%lld", &n,&t);
        for (int i=1;i<=n;i++) scanf ("%d", &val[i]);
        Sort (val+1,val+1+n);
        int k=0;
        int l=1,r=n;
            while (L<r) {k= (l+r)/2;
            if (merge (k)) r=k;
        else l=k+1;
    } printf ("%d\n", L);
} return 0; }