POJ-3253--Fence Repair(简单哈夫曼的应用)

Akatsuki

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 Li (1 ≤ Li ≤ 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 Li). 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
3
8
5
8
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 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).

题意：（直接讲样例）给一个数3，然后给了3个木板（已经切割好的），加起来是21。第一次将它切为两半变为 8和13，费用是8+13=21；第二次选取16那一块将它切为两半为 5和8，此时费用为21+5+8=34.当然了切割方式有很多种，这里是让你找最优的切割方式来使所用的费用最小。

思路：要使费用达到最小，就要让每次切割产生的那个下次不被切割的木板尽可能的大，由于由整到分找寻最小花费不好找，我们便想到由给出的数据逆回去直到组合成完整的木板，每次都选取两个最小的来组合并以此类推（哈弗曼思想），直到组合成一整个木板，此时的话费就是最小花费。
例：3 4 5 6。先选3和4，此时花费为7，然后再选5和6，此时花费为7+11=18，再选 7和11，此时花费为18+7+11=36.最小花费为36元。

#include<iostream>#include<queue>#include<algorithm>using namespace std;int main(){    int n;    cin>>n;    int i;    priority_queue<int,vector<int>,greater<int> >q;//建一个从小到大的优先队列    long long int sum=0;    for(i=0;i<=n-1;i++)    {        int m;        cin>>m;        q.push(m);    }    while(!q.empty())    {        int x,y;        x=q.top();        q.pop();        y=q.top();        q.pop();        sum=sum+x+y;        if(q.empty())            break;//这里要注意，因为下面每次都会入队一个元素，所以要提前判断队列是否为空，否则会死循环。        q.push(x+y);    }    cout<<sum<<endl;}