pku2363 Yogurt factory
每周要供应一定数目的奶酪，但是每周的生产成本可能会变化，而存储成本不会变化，问如何生产代价最低。
Sample Input
4 5 88 200 89 400 97 300 91 500
Sample Output
126900
Hint
OUTPUT DETAILS:
In week 1, produce 200 units of yogurt and deliver all of it. In week 2, produce 700 units: deliver 400 units while storing 300 units. In week 3, deliver the 300 units that were stored. In week 4, produce and deliver 500 units.
贪心问题，有可能某周多生产，下周不生产，因为下周的生产成本太大，以至于宁可多存储一天，甚至两天，三天....。
事实上某周多生产，可以还是看成是在下一周生产，只是我们用等价生产成本来看，
本周等价生产成本 = min(上一周的等价生产成本+存储一周的成本, 这周的生产成本)
如果
存储成本为5
第一周生产成本80，第二周86，第三周92
那么其实应该第一周生产完这三周的供应量。相当与第一周等价生产成本80，第二周等价生产成本85，第三周90.随着输入，扫描处理一遍即可得到最优解。
不过下面的程序比较耗时，似乎是用cout,cin的关系，用printf,scanf会快很多。
1
#include
<
iostream
>
2 using namespace std;
3
4 void GetMinCost()
5 {
6 long long cost = 0 ;
7 int weekNum, makeCost, storeCost, supplyNum;
8 cin >> weekNum >> storeCost;
9 int preMakeCost, minCost;
10 // the first week
11 cin >> makeCost >> supplyNum;
12 cost += makeCost * supplyNum;
13 preMakeCost = makeCost + storeCost;
14 // other weeks
15 for ( int i = 1 ; i < weekNum; i ++ ) {
16 cin >> makeCost >> supplyNum;
17 minCost = min(preMakeCost, makeCost);
18 cost += minCost * supplyNum;
19 preMakeCost = minCost + storeCost;
20 }
21
22 cout << cost << endl;
23 }
24
25 int main( int argc, char * argv[])
26 {
27 GetMinCost();
28 return 0 ;
29 }
2 using namespace std;
3
4 void GetMinCost()
5 {
6 long long cost = 0 ;
7 int weekNum, makeCost, storeCost, supplyNum;
8 cin >> weekNum >> storeCost;
9 int preMakeCost, minCost;
10 // the first week
11 cin >> makeCost >> supplyNum;
12 cost += makeCost * supplyNum;
13 preMakeCost = makeCost + storeCost;
14 // other weeks
15 for ( int i = 1 ; i < weekNum; i ++ ) {
16 cin >> makeCost >> supplyNum;
17 minCost = min(preMakeCost, makeCost);
18 cost += minCost * supplyNum;
19 preMakeCost = minCost + storeCost;
20 }
21
22 cout << cost << endl;
23 }
24
25 int main( int argc, char * argv[])
26 {
27 GetMinCost();
28 return 0 ;
29 }