[ACM] poj 3253 fence repair (Huffman tree idea, priority queue)

Fence repair

Time limit:2000 ms		Memory limit:65536 K
Total submissions:25274		Accepted:8131

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds that he needsN(1 ≤N≤ 20,000) planks of wood, each having some integer LengthLi(1 ≤Li≤ 50,000) units. He then purchases a single long board just long enough to saw intoNPlanks (I. e., whose length is the sum of the lengthsLi). Fj is ignoring the "kerf", the extra length lost to sawdust when a sawcut is made; You shoshould 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 ofN-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 createNPlanks. fj knows that he can cut the board in varous 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

3858

Sample output

34

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 shoshould be used to cut the Board into pieces measuring 13 and 8. the second cut will cost 13, and shoshould be used to cut the 13 into 8 and 5. this wocould cost 21 + 13 = 34. if the 21 was cut into 16 and 5 instead, the second cut wocould cost 16 for a total of 37 (which is more than 34 ).

Source

Usaco 2006 November gold

Solution:

Construct a Huffman tree for the input data. The sum of the values of non-leaf nodes is required.

The construction process of the Huffman tree: Get the two least numbers in all numbers each time and add them together to get the value of a non-leaf node. The two least numbers obtained before next time cannot be obtained, then, the values of non-leaf nodes can be obtained.

For example, 3, 4, 5, add 3, 4 to 7, and then take 5, 7 (3, 4 can no longer be obtained), and add 12, 7 + 12 = 19, that is, the request.

This operation can be well implemented by the priority queue.

Code that defines the int type priority from small to large:

priority_queue<int ,vector<int>,greater<int> > q

Code:

#include <iostream>#include <queue>using namespace std;int main(){    priority_queue<int ,vector<int>,greater<int> > q;    int n;    int num;    cin>>n;    while(n--)    {        cin>>num;        q.push(num);    }    long long sum=0;        while(!q.empty())    {        int n1=q.top();        q.pop();        int n2=q.top();        q.pop();        sum+=n1+n2;        if(q.empty())            break;        q.push(n1+n2);    }        cout<<sum<<endl;    return 0;}