HDU 5884 Sort ——（K叉哈夫曼树）

 1 #include <stdio.h>
 2 #include <algorithm>
 3 #include <string.h>
 4 #include <queue>
 5 using namespace std;
 6 const int N = 100000 + 5;
 7 typedef long long ll;
 8 
 9 int n,lim;
10 int a[N];
11 
12 bool can(int k)
13 {
14     if(k == 1) return false;
15     queue<ll> Q1,Q2;
16     ll sum = 0;
17     for(int i=1;i<=n;i++) Q1.push((ll)a[i]);
18     while(Q1.size()+Q2.size() > 1)
19     {
20         if(Q1.size()+Q2.size() >= k)
21         {
22             ll temp = 0;
23             for(int i=1;i<=k;i++)
24             {
25                 if(Q2.size()==0)
26                 {
27                     temp += Q1.front();Q1.pop();
28                 }
29                 else if(Q1.size()==0)
30                 {
31                     temp += Q2.front();Q2.pop();
32                 }
33                 else if(Q1.front()<=Q2.front())
34                 {
35                     temp += Q1.front();Q1.pop();
36                 }
37                 else
38                 {
39                     temp += Q2.front();Q2.pop();
40                 }
41             }
42             sum += temp;
43             Q2.push(temp);
44         }
45         else
46         {
47             ll temp = 0;
48             while(!Q1.empty())
49             {
50                 temp += Q1.front();Q1.pop();
51             }
52             while(!Q2.empty())
53             {
54                 temp += Q2.front();Q2.pop();
55             }
56             sum += temp;
57         }
58     }
59     return sum <= (ll)lim;
60 }
61 
62 int main()
63 {
64     int T;scanf("%d",&T);
65     while(T--)
66     {
67         scanf("%d%d",&n,&lim);
68         for(int i=1;i<=n;i++) scanf("%d",a+i);
69         sort(a+1,a+1+n);
70         int L = 2, R = n;
71         int ans = 1;
72         while(L <= R)
73         {
74             //printf("!! %d %d\n",L,R);
75             int mid = L + R >> 1;
76             if(can(mid))
77             {
78                 ans = mid;
79                 R = mid - 1;
80             }
81             else L = mid + 1;
82         }
83         printf("%d\n",ans);
84     }
85     return 0;
86 }
87 
88 /*
89     6
90     6   120
91     10 10 10 10 10
92 */

 1 #include<iostream>
 2 //#include<bits/stdc++.h>
 3 #include<cstdio>
 4 #include<string>
 5 #include<cstring>
 6 #include<map>
 7 #include<queue>
 8 #include<set>
 9 #include<stack>
10 #include<ctime>
11 #include<algorithm>
12 #include<cmath>
13 #include<vector>
14 #define showtime fprintf(stderr,"time = %.15f\n",clock() / (double)CLOCKS_PER_SEC)
15 #pragma comment(linker, "/STACK:1024000000,1024000000")
16 using namespace std;
17 typedef long long ll;
18 typedef long long LL;
19 #define MP make_pair
20 #define PII pair<int,int>
21 #define PLI pair<long long ,int> 
22 #define PFI pair<double,int>
23 #define PLL pair<ll,ll>
24 #define PB push_back
25 #define F first
26 #define S second
27 #define lson l,mid,rt<<1
28 #define rson mid+1,r,rt<<1|1
29 #define debug cout<<"?????"<<endl;
30 //freopen("1005.in","r",stdin);
31 //freopen("data.out","w",stdout);
32 const int INF = 0x7f7f7f7f;
33 const double eps = 1e-2;
34 const int M = 1000000  + 50;
35 const int N = 1000 + 50 ;
36 const double PI = acos(-1.);
37 const double E = 2.71828182845904523536;
38 const int MOD = 100000007;
39 typedef vector<ll> Vec;
40 typedef vector<Vec> Mat;
41 int T,n,a[100000 + 50];
42 ll m;
43 bool ok(int k){
44     queue<ll> q,p;
45     int t = (n-1) % (k-1);  
46     // 每次减少k-1个数。一共要减少 (n-1) 个数。 还剩下几个数要和 0 一组了
47     if(t != 0) for(int i = 0 ; i < k - t - 1 ; i ++) q.push(0);
48     for(int i = 0 ; i < n ; i ++) q.push(a[i]);
49 
50     ll ans = 0;
51     while(!q.empty() || !p.empty()){
52         ll tmp = 0;
53         for(int i = 0 ; i < k ; i ++){
54             if(!q.empty() && !p.empty()){
55                 ll u = q.front() , v = p.front();
56                 if(u < v) tmp += u , q.pop();
57                 else tmp += v , p.pop();
58             }else if(!q.empty()){
59                 ll u = q.front(); q.pop();
60                 tmp += u;
61             }else if(!p.empty()){
62                 ll v = p.front() ; p.pop();
63                 tmp += v;
64             }else break;
65         }
66         ans += tmp;
67         if(q.empty() && p.empty()) break;
68         p.push(tmp);
69     }
70     return ans <= m;
71 }
72 void solve(){
73     int l = 2 , r = n;
74     while(l < r){
75         int m = (l+r)>>1;
76         if(ok(m)) r = m;
77         else l = m + 1;
78     }
79     cout << l << endl;
80 }
81 int main(){
82     cin >> T;
83     while(T --){
84         cin >> n >> m;
85         for(int i = 0 ; i < n ; i ++) scanf("%d",&a[i]);
86         sort(a,a+n);
87         solve();
88     }
89     return 0;
90 }