贪心算法HURUST题目
题目描述:
Yogurt factory
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) cents to produce one unit of yogurt in week i. Yucky's factory, being well-designed, can produce arbitrarily many units of yogurt each week.
Yucky Yogurt owns a warehouse that can store unused yogurt at
a constant fee of S (1 <= S <= 100) cents per unit of yogurt per
week. Fortuitously, yogurt does not spoil. Yucky Yogurt's warehouse is
enormous, so it can hold arbitrarily many units of yogurt.
Yucky wants to find a way to make weekly deliveries of Y_i (0
<= Y_i <= 10,000) units of yogurt to its clientele (Y_i is the
delivery quantity in week i). Help Yucky minimize its costs over the
entire N-week period. Yogurt produced in week i, as well as any yogurt
already in storage, can be used to meet Yucky's demand for that week.
* Lines 2..N+1: Line i+1 contains two space-separated integers: C_i and Y_i.
4 5 88 200 89 400 97 300 91 500
126900
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这道题是我在HURBST做的题目,链接是https://vjudge.net/contest/262212#problem/D
中文理解:有一个酸奶酪生产商,每一个星期都有其需求量和价格量,奶酪可以保存到下个星期使用,不过有保存费用,此题目要求你找出最小费用的解;
个人看法:这道题目是贪心的基础题目,它满足了贪心算法的基于当前是最优解,找出下一个最优解,保证这一步是最优解。
1 #include<stdio.h> 2 #include<algorithm> 3 #include<math.h> 4 #include<vector> 5 #include<iterator> 6 #include<string> 7 #include<string.h> 8 #include<ctype.h> 9 #include<map> 10 #include<stack> 11 #include<queue> 12 #include<iostream> 13 #include<time.h> 14 15 using namespace std; 16 17 #define rep(i ,a, b) for(int i = a; i <= b; i++) 18 #define per(i, a, b) for(int i = a; i <= b; i--) 19 typedef long long ll;//数据量大,用longlong来进行运算较合适。不然会数据溢出。 20 21 struct milk 22 { 23 ll c_i, y_i, id; 24 }a[10006]; 25 26 ll findmin(ll n, ll s) 27 { 28 ll ans=a[0].c_i*a[0].y_i, k = 0; 29 for(ll i = 1; i < n; i++) 30 { 31 ll temp = a[k].c_i+s*(i- a[k].id), cur = a[i].c_i;//记录前一周牛奶的价格和这个一周的价格,进行比较然后,选取较小的作为当前的最小解; 32 if(temp < cur) 33 { 34 ans += temp*a[i].y_i; 35 } 36 else 37 { 38 ans += cur*a[i].y_i; 39 k = i; 40 } 41 } 42 return ans; 43 } 44 45 int main()//这是贪心问题。 46 { 47 ll n, s; 48 while(scanf("%lld%lld", &n, &s) != EOF) 49 { 50 for(ll i =0; i < n; i++) 51 { 52 scanf("%lld%lld", &a[i].c_i, &a[i].y_i); 53 a[i].id = i; 54 } 55 printf("%lld\n", findmin(n, s)); 56 } 57 return 0; 58 }