POJ 3253 - Fence Repair（贪心）_armer john wants to repair a small length of the f-优快云博客

Fence Repair

Time Limit: 2000MS		Memory Limit: 65536K
Total Submissions: 21562		Accepted: 6886

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

Sample Output

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).

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看Hint就知道这题是什么意思了

最开始的思路是从大到小排个序，花费从总数开始，每次减去之前砍掉的部分的长度，一发WA发现问题。

给一组数据：4 2 2 3 3，按我之前的做法得到21，但实际结果是20。先切成4 6，再分别切2 2和3 3。

发现好像是哈夫曼树，从来没写过参考了下别人的代码，是用优先队列写的

顺便学会了优先队列直接int类型从小到大排列的方法：priority_queue< int,vector<int>,greater<int> >que;

注意答案用long long

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <queue>

using namespace std;

int main()
{
    int n,a[22222];
    scanf("%d",&n);
    long long ans=0;
    priority_queue< int,vector<int>,greater<int> >que;//从小到大。从大到小用less<int>，不过从大到小的话直接用priority_queue<int>que;默认的就行了
    for(int i=0;i<n;i++)
    {
        scanf("%d",&a[i]);
        que.push(a[i]);
    }
    for(int i=0;i<n-1;i++)
    {
        int a=que.top(); que.pop();
        int b=que.top(); que.pop();
        //cout<<"a="<<a<<" b="<<b<<endl;
        ans+=(a+b);
        que.push(a+b);
    }
    printf("%I64d\n",ans);
    return 0;
}