POJ - 3253 Fence Repair 优先队列+贪心

Fence Repair

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 Nplanks. 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).

 

 

优先队列，起初考虑每次切掉最大的边，但不一定每次都从边上切，有先切中间再切两边的情况，如12 8 13 7就是40 27 15而正解是40 20 20显然错误，所以这种贪心行不通。逆思维，可以把切的过程倒过来想，每次切开的一定是最小边，所以每次找到两个最小边拼在一起，计算拼接长度和即可。

 

#include<stdio.h>
#include<queue>
using namespace std;

int main()
{
    int n,x,a,b,i;
    long long min;
    priority_queue<int,vector<int>,greater<int> >q;
    scanf("%d",&n);
    for(i=;i<=n;i++){
        scanf("%d",&x);
        q.push(x);
    }
    min=;
    while(q.size()>){
        a=q.top();q.pop();
        b=q.top();q.pop();
        min+=a+b;
        q.push(a+b);
    }
    printf("%lld\n",min);
    return ;
}