HDU 5350(MZL's munhaff function-哈夫曼树)

MZL's munhaff function

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 230    Accepted Submission(s): 133

Problem Description

MZL is a mysterious mathematician, and he proposed a mysterious function at his young age.
Stilwell is very confused about this function, and he need your help.
First of all, given n positive integers Ai and Ai≥Ai+1.
Then, generate n positive integers Bi

Bi=∑j=inAj

Define f(i,j) for i,j∈Z

f(i,j)=⎧⎩⎨⎪⎪⎪⎪⎪⎪0min(f(i−1,j+1),f(i,⌈j2⌉)+Bi)1011037(i,j)=(1,1)i,j∈[1,n], (i,j)≠(1,1)otherwise

Find f(n,1).

 

Input

The first line of the input contains a single number T, the number of test cases.
For each test case, the first line contains a positive integer n, and the next line contains n positive integers Ai.
T≤100,1≤n≤105,∑n≤106,1≤Ai≤104.

 

Output

For each test case, output f(n,1) in a line.

 

Sample Input

331 1 1528 26 25 24 110996 901 413 331 259 241 226 209 139 49

 

Sample Output

523311037
Hint
case 1 ：f(1,1)=0f(1,2)=f(1,1)+3=3f(1,3)=f(1,2)+3=6f(2,1)=min(f(2,1)+2,f(1,2))=3f(2,2)=min(f(2,1)+2,f(1,3))=5f(2,3)=f(2,2)+2=7f(3,1)=min(f(3,1)+1,f(2,2))=5
 

 

Author

SXYZ

 

Source

2015 Multi-University Training Contest 5

 

Recommend

wange2014   |   We have carefully selected several similar problems for you:  5421 5420 5419 5418 5417 

 

哈夫曼树

解决如下问题：有一堆数，每次拿2个数，换成它们的和，代价是它们的和，求把数变为1个的最小花费

显然每次取当前最小（构建哈夫曼树）

考虑本题，f[i][j] 表示拿了 前Ai个数，当前最后那层有j个空的叶节点

我们可以拿一个数去填空点，不需要代价f[i+1][j-1] 

也可以把叶子节点上有数往下画一层 f[i][2j] 代价就是上面所有取的数 Bi 

#include<bits/stdc++.h> using namespace std;#define For(i,n) for(int i=1;i<=n;i++)#define Fork(i,k,n) for(int i=k;i<=n;i++)#define Rep(i,n) for(int i=0;i<n;i++)#define ForD(i,n) for(int i=n;i;i--)#define RepD(i,n) for(int i=n;i>=0;i--)#define Forp(x) for(int p=pre[x];p;p=next[p])#define Forpiter(x) for(int &p=iter[x];p;p=next[p])  #define Lson (x<<1)#define Rson ((x<<1)+1)#define MEM(a) memset(a,0,sizeof(a));#define MEMI(a) memset(a,127,sizeof(a));#define MEMi(a) memset(a,128,sizeof(a));#define INF (2139062143)#define F (100000007)#define MAXN (1000000+10)#define MAXAi (10000+10)typedef long long ll;ll mul(ll a,ll b){return (a*b)%F;}ll add(ll a,ll b){return (a+b)%F;}ll sub(ll a,ll b){return (a-b+llabs(a-b)/F*F+F)%F;}void upd(ll &a,ll b){a=(a%F+b%F)%F;}int n;ll a[MAXN];priority_queue<ll,vector<ll> , greater<ll> > q;int main(){//freopen("H.in","r",stdin);int T;cin>>T;while (T--) {cin>>n;For(i,n) {scanf("%lld",&a[i]);q.push(a[i]);}////if (n==1) {//puts("0");continue;//}ll ans=0;while (q.size()>=2){ll val=q.top(); q.pop();val+=q.top(); q.pop();ans+=val;q.push(val);}cout<<ans<<endl;while (!q.empty()) q.pop();}return 0;}