poj 3253 Fence Repair（贪心+优先队列）

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 36179		Accepted: 11729

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

题目大意：

给定一个木条，锯成input中要求的段数和每一段的长度，每锯一次花费锯成被锯成两半的木条长度之和，求最小花费

解题思路：

先把题目转化为把一些小木条合并成一个大木条，这是一个贪心的题目，实际上就是霍夫曼编码的原理，每次都合并当前最小的两个木条。求最小的时候用优先队列可以优化到nlogn

ps：没有用long long输出ansWA了一发，因为数据规模是L*N*N的，误以为是L*N了。

#include<stdio.h>
#include<queue>
#define ll long long
using namespace std;
priority_queue<int, vector<int>, greater<int> > q;//小顶堆
int main()
{
    int n;
    ll ans=0;
    scanf("%d",&n);
    for(int i=0;i<n;i++)
    {
        int tem;
        scanf("%d",&tem);
        q.push(tem);
    }
    while(1)
    {
        int tem1=q.top();
        q.pop();
        if(q.empty())break;
        int tem2=q.top();
        q.pop();
        ans+=(ll)tem1+tem2;
        q.push(tem1+tem2);
    }
    printf("%lld\n",ans);
}