POJ 3253 Fence Repair

Fence Repair

Time Limit: 2000MS Memory Limit: 65536KTotal Submissions: 25668 Accepted: 8263

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 lengthL_i (1 ≤ L_i ≤ 50,000) units. He then purchases a single long board just long enough to saw into theN 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 theN-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 theN 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 makeN-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 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：
有一根木板，切成 n 块，已知最终每块木板长度，求最小花费。每次花费为本次所切木板长度。
分析：
最小花费即每次都切成两个最小长度，例如有长度分别为 6,5,7,4,2 的木板，则最后一次切的长度为 6，切成了4,2。本次未切之前木板为6,5,7，6，同理，倒数第二次未切之前为
11,7,6.倒数第三次未切之前为11,13.总长为24.共花费54.
用优先队列保存每块木板长度，每次从头部取两个最小长度，求和，累加，再加入队列。。。
代码如下：

#include<cstdio>#include<iostream>#include<queue>using namespace std;int main(){    priority_queue<int, vector<int>, greater<int> > q;  //由小到大的优先队列，头部最小    int n, m;    while(cin >>n)    {        while(!q.empty())  //清空队列            q.pop();        while(n--)        {            scanf("%d", &m);            q.push(m);  //加入队列        }        long long s = 0;  //数据太大，int就WA了        while(q.size() > 1)        {            int k = q.top();  //先从头部取一个            q.pop();  //除去头部元素            k += q.top();  //再从头部取一个，次头为上一个头下面的元素            q.pop();            s += (long long)k;            q.push(k);   //将两最小值的和加入队列        }        cout <<s <<endl;    }    return 0;}