ACM_贪心法_queue_Fence Repair

题目如下，来源POJ

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

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).

代码如下：

#include <stdio.h>
#include <queue>
using namespace std;
priority_queue<long long int,vector<long long int>,greater<long long int> >p;
long long int n,b,c,d,cou;
long long int plank;
int main(){while(~scanf("%d",&n)&&n){cou=0;d=0;for(int a=0;a<n;a++){scanf("%d",&plank);p.push(plank);}while(1){b=0;c=0;b=p.top();p.pop();if(p.empty()) break;c=p.top();p.pop();d=b+c;cou=d+cou;p.push(d);}printf("%lld\n",cou);}return 0;
}

　　这题一眼看去，第一思路是由整块木板开始，先把大块的切下来，但这种思路是错误的（可证反）。正确思路是从树表底层开始，不断将最小的两块合并，提取每次合并的和，即可得到答案，常用于霍夫曼编码规则（从树状表底层开始编码）。用queue更易解。

转载于:https://www.cnblogs.com/Never-Land/p/10103911.html

ACM_贪心法_queue_Fence Repair相关推荐

《c陷阱与缺陷》之贪心法
在词法分析中,有条规则:每个符号应该包含尽可能多的字符,被称为"贪心法"或"大嘴法". K&R表述如下:如果(编译器的)输入流截止至某个字符之前都已经被 ...
贪心问题JAVA_java背包问题(贪心法)
完全背包问题一个旅行者有一个最多能用m公斤的背包,现在有n种物品,每件的重量分别是W1,W2,...,Wn,每件的价值分别为C1,C2,...,Cn.若的每种物品的件数足够多.求旅行者能获得的最大总 ...
【LeetCode笔记】55. 跳跃游戏（Java、贪心法）
文章目录题目描述解法 & 思路题目描述可以用动态规划来做,dp[i]代表以第i个下标做起点,可以到达的最远的地方,从后往前走,最后dp[0] >= len-1说明可达. 不过我的 ...
1214线段覆盖问题——贪心法
题目描述: 给定x轴上的N(0<N<100)条线段,每个线段由它的二个端点a_I和b_I确定,I=1,2,--N.这些坐标都是区间(-999,999)的整数.有些线段之间会相互交叠或覆盖. ...
学渣的刷题之旅 leetcode刷题 53.最大子序和（动态规划，贪心法）
给定一个整数数组 nums ,找到一个具有最大和的连续子数组(子数组最少包含一个元素),返回其最大和. 输入: [-2,1,-3,4,-1,2,1,-5,4], 输出: 6 解释: 连续子数组 [4, ...
【算法】一文详解贪心法
一.概述贪心法将一个复杂问题分解为一系列较为简单的局部最优解,每一步都是对当前解的一个扩展,直到获得问题的完全解.贪心法的典型应用时求解最优化问题,而且即使是非最优解,最终得出的解也和最优解比较近似 ...
NOJ 1056 地道普里姆算法+贪心法
复习算法至普里姆算法,书上的代码不全,在oj上找了一题巩固一下. 普里姆算法应用在最小代价生成树一类的题型中,其主要执行思想是:选择一个点作为源点,此时生成树只包含一个源点,后从未加入生成树的点中选择 ...
算法设计与分析-----贪心法
算法设计与分析-----贪心法(c语言) 一.贪心法 1.定义 2.贪心法具有的性质 1.贪心选择性质 2.最优子结构性质 3.贪心法的算法框架 5.求解活动安排问题 6.求解最优装载问题二. 贪心 ...
Java乘船_ACM：贪心法：乘船问题。
题目:有n个人,第i个人的重量为wi,每艘船的最大载重量均为C,且最多只能乘两个人.用最少的船装载所有人. 分析:贪心法! 考虑最轻的人i,他应该和谁一起坐呢?如果每个人都无法和他一起坐船,那么唯一的 ...

ACM_贪心法_queue_Fence Repair

ACM_贪心法_queue_Fence Repair相关推荐

最新文章

热门文章