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Yogurt factory
一道简单的贪心模板题 http://poj.org/problem?id=2393 将每天的价格都放到最低就行了 核心思路：第i天最低的价格 = min(第i-1天最低的价格+s,第i天原本的价格) 有了思路，代码随便打 #include <iostream> #include <algorithm> using namespace std; #define maxn 10005 typedef loYogurt factory POJ 2393(贪心)
原题 题目链接 题目分析 简单贪心题,从第一周开始,每过一周就把当周酸奶的生产价格加入优先队列中,不过由于存储要钱,所以加入优先队列需要一点额外操作.想一下当i周时,第一周生产的酸奶加上存储成本c1=c1+s*(i-1),第二周生产的酸奶同理,c2=c2+s*(i-2)...我们要从中选处理后价格最Yogurt factory POJ 2393(贪心)
原题 题目链接 题目分析 简单贪心题,从第一周开始,每过一周就把当周酸奶的生产价格加入优先队列中,不过由于存储要钱,所以加入优先队列需要一点额外操作.想一下当i周时,第一周生产的酸奶加上存储成本c1=c1+s*(i-1),第二周生产的酸奶同理,c2=c2+s*(i-2)...我们要从中选处理后价格最BZOJ 1740: [Usaco2005 mar]Yogurt factory 奶酪工厂 贪心_问题转化
好久以前碰到的题，真的好巧妙 Code: #include<bits/stdc++.h>#define setIO(s) freopen(s".in","r",stdin) #define ll long long using namespace std;int main(){ // setIO("input"); int n; ll s,c,y,cur=0,ans=0; scanf("%d%lld"POJ 2393 Yogurt factory
Yogurt factory Time Limit: 1000MS Memory Limit: 65536K Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such thatPOJ2393 Yogurt factory （贪心）
Description The cows have purchased a yogurt factory that makes world-famous Yucky Yogurt. Over the next N (1 <= N <= 10,000) weeks, the price of milk and labor will fluctuate weekly such that it will cost the company C_i (1 <= C_i <= 5,000) c