POJ 3253 Fence Repair: Greed and Priority queues

Fence Repair

http://poj.org/problem?id=3253

Time limit:2000ms

Memory limit:65536k

Description

Farmer John wants to repair a small length of the fence around the pasture. He measures the fence and finds, "he needs n " (1≤ n ≤20,000) Planks of wood, each having some intege R 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 him doesn ' t own a saw with which to cut the wood and so he mosies over to farmer Don ' s Farm with this L Ong 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 was exactly equal to its length. Cutting a plank of length costs cents.

Farmer Don then lets farmer John decide the order and locations to cut the plank. Help farmer John Determine the minimum amount of the $ he can spend to create the N planks. FJ knows and can cut the board in various different orders which'll 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 $ 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 into pieces of lengths 8, 5, and 8.
The original board measures 8+5+8=21. The "a" and should is used to cut the board into pieces measuring and 8. The second cut would cost, and should is used to cut the 8 and 5. This is would cost 21+13=34. If the second cut would the $ for a total of 5 instead (which are more than 34).

Source

Usaco 2006 November Gold

Water over ~

See more highlights of this column: http://www.bianceng.cnhttp://www.bianceng.cn/Programming/sjjg/

Complete code:

/*32ms,480kb*/
  
#include <cstdio>
#include <queue>
using namespace std;
typedef long long LL;
  
Priority_queue<ll, Vector<ll>, greater<ll> > pq;///ascending
  
int main (void)
{
    int n,len;
    ll ans;
    while (~SCANF ("%d", &n))
    {
        if (!pq.empty ()) Pq.pop ();
        for (int i = 0; i < N; ++i)
            scanf ("%d", &len), Pq.push (len);
        Ans = 0;
        while (Pq.size () > 1)
        {
            ll a = Pq.top ();
            Pq.pop ();
            ll B = Pq.top ();
            Pq.pop ();
            Ans + A + b;
            Pq.push (A + b);
        }
        printf ("%lld\n", ans);
    }
    return 0;
}