POJ 3253 Fence Repair（哈夫曼树）

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 26167		Accepted: 8459

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needs N (1 ≤ N ≤ 20,000) planks of wood, each having some integer length L_i (1 ≤ L_i ≤
50,000) units. He then purchases a single long board just long enough to saw into the N planks (i.e., whose length is the sum of the lengths L_i). FJ is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made;
you should ignore it, too.

FJ sadly realizes that he doesn't own a saw with which to cut the wood, so he mosies over to Farmer Don's Farm with this long board and politely asks if he may borrow a saw.

Farmer Don, a closet capitalist, doesn't lend FJ a saw but instead offers to charge Farmer John for each of the N-1 cuts in the plank. The charge to cut a piece of wood is exactly equal to its length. Cutting a plank of length 21 costs 21 cents.

Farmer Don then lets Farmer John decide the order and locations to cut the plank. Help Farmer John determine the minimum amount of money he can spend to create the N planks. FJ knows that he can cut the board in various different orders which will
result in different charges since the resulting intermediate planks are of different lengths.

Input

Line 1: One integer N, the number of planks

Lines 2..N+1: Each line contains a single integer describing the length of a needed plank

Output

Line 1: One integer: the minimum amount of money he must spend to make N-1 cuts

Sample Input

Sample Output

Hint

He wants to cut a board of length 21 into pieces of lengths 8, 5, and 8.

The original board measures 8+5+8=21. The first cut will cost 21, and should be used to cut the board into pieces measuring 13 and 8. The second cut will cost 13, and should be used to cut the 13 into 8 and 5. This would cost 21+13=34. If the 21 was cut into
16 and 5 instead, the second cut would cost 16 for a total of 37 (which is more than 34).

Source

USACO 2006 November Gold

#include <iostream>

using namespace std;

#define maxn 20010

long long int n;//目标板数

long long int len;//堆长

long long int p[maxn];//堆

void swap(long long int &a,long long int &b)

{

    long long int temp;

    temp=a;

    a=b;

    b=temp;

}

void heap_insert(long long int k)//将k插入到小根堆中。并维护堆性质

{

    long long int t=++len;//将k插入到堆尾

    p[t]=k;

    while(t>1)//自上而下将k调整到合适的位置

    {

        if(p[t/2]>p[t])//若t的值大于其父节点的值，则交换。继续向上调整

        {

            swap(p[t/2],p[t]);

            t=t/2;

        }

        else

            break;

    }

}

void heap_pop()

{

    long long int t=1;//将堆尾元素移到堆首

    p[t]=p[len--];

    while(2*t<=len)//调整堆首元素到合适位置，自上而下调整

    {

        long long int k=2*t;

        if(k<len&&p[k]>p[k+1])//计算左右儿子中较小的节点序号k

            k=k+1;

        if(p[t]>p[k])

        {

            swap(p[t],p[k]);

            t=k;

        }

        else

            break;

    }

}

int main()

{

    cin>>n;

    long long int i;

    for(i=1;i<=n;i++)

        cin>>p[i];

    for(i=1;i<=n;i++)

        heap_insert(p[i]);//将n块木板的长度增加小根堆

    long long int ans;//最小费用

    ans=0;

    while(len>1)//构造哈夫曼树

    {

        long long int a,b;

        a=p[1];//取堆首节点（权值a）。并维护其性质

        heap_pop();

        b=p[1];

        heap_pop();

        ans=ans+a+b;//将a和b累计计入最小费用中

        heap_insert(a+b);//合并成一个权值插入到小根堆

    }

    cout <<ans<<endl;

    return 0;

}

巴特西

POJ 3253 Fence Repair（哈夫曼树）

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