poj 3053 优先队列处理

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 39029		Accepted: 12681

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 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 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).

Source

USACO 2006 November Gold

 

题意： 切木板问题 给出你  要求的n块木板的大小  并且定义每次切板 的代价数值上等于原木板的大小

          为完成要求 输出代价最小值。

 

题解： 每次选取最少的两个合并，直到剩一个。

         然而这样，每次都要排序以确定最少两个木板，代码复杂度、时间复杂度和空间复杂度都要上升

        STL优先队列处理 将 要求的n块木板入队，每次将最小的两个s1，s2 出队    求和s1+s2并累加到sum 说明s1+s2大小的木板是 s1，s2的原板

        取出最小的两个是为了满足  用最小的代价得到 当前最小的两块木板  之后将s1+s2入队  直到队列中只剩下 一个元素

        输出sum

 

#include<iostream>
#include<cstring>
#include<algorithm>
#include<cstdio>
#include<queue>
#define ll long long
#define mod 1e9+7
#define PI acos(-1.0)
#define bug(x) printf("%%%%%%%%%%%%%",x);
#define inf 1e8
using namespace std;
priority_queue<int, vector<int>, greater<int> > q;
int n;
int a;
ll ans=0;
int main()
{
    while(scanf("%d",&n)!=EOF)
{
    ans=0;
    for(ll i=1;i<=n;i++)
    {
        scanf("%d",&a);
        q.push(a);
    }
    ll s1,s2;
    while(q.size()>1)
    {
        s1=q.top();
        q.pop();
        s2=q.top();
        q.pop();
        ans=ans+s1+s2;
        q.push(s1+s2);
    }
    q.pop();
    printf("%lld
",ans);
}

    return 0;
}