Hdu 5350 MZL's munhaff function 2015ACM多校对抗赛第五场

传送门： http://acm.hdu.edu.cn/showproblem.php?pid=5350

MZL's munhaff function

Time Limit: 3000/1500 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 156    Accepted Submission(s): 101

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
 

 

Source

2015 Multi-University Training Contest 5

 

神奇的Huffman树。

Ai是Huffman的叶子节点，将A降序排列

dp[i][j]表示放置Ai，还剩j个叶子节点，Huffman的取值。

转移有两种情况：

放在叶子节点处，dp[i+1][j-1]，叶子节点还剩j-1个，Ai已经放置好了，考虑Ai+1。

放在非叶子节点处，即下一层，dp[i][j*2]，还要考虑次Ai，但是叶子节点扩展两倍了。

和题目给的递推正好互逆。

考虑Bi的值，发现，最后答案f[n][1]就是Huffman树的权值和 减去 叶子节点权值和。

#include<iostream>#include<cstdio>#include<queue>#include<algorithm>using namespace std;#define LL long longstruct node{    LL val;    node(LL v=0):val(v){}    bool operator < (const node &a)const{        return val>a.val;    }};priority_queue <node> que;int main(){    LL T,n;scanf("%lld",&T);    while(T--){        scanf("%lld",&n);        while(!que.empty())que.pop();        for(int i=0;i<n;i++){            LL val;scanf("%lld",&val);            que.push(node(val));        }        if(n==1){puts("0");continue;}        LL ans=0;        while(que.size()>=2){            LL val=que.top().val;que.pop();              val+=que.top().val;que.pop();            ans+=val;            que.push(node(val));        }        printf("%lld\n",ans);    }    return 0;}