贪心算法之Air Express——by sicily
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写在前面
有时候我经常想
为什么每次机考我会花超长的时间来A一道题
直到最近我才明白
原来我每次机考的时候
不是打代码
而是在花时间改代码
思路并不难
但是时间往往会花的超长
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Sicily1093 Air Express
Description
Fly It Today! (FIT), an air express company, charges different amounts for packages depending on their weight. For example, one set of rates may be:
Package weight Cost per pound
0 to 9 pounds $10
10 to 49 pounds $5
50 to 99 pounds $3
100 pounds or more $2
This rate structure has upset some customers who have realized that it costs less to ship a 10pound package ($50) than an 8 pound package ($80) and it costs less to ship a 100 poundpackage ($200) than a 90 pound one ($270). FIT wants to check packages to determine if the customer can pay a lower price by adding weight to the package. If this is the case, they want to know the minimum weight to be added to obtain the lowest price possible.
Input
The input file will have one or more data sets. Each data set begins with exactly 4 lines, giving the shipping rates. These will be:
weight1 rate1
weight2 rate2
weight3 rate3
rate4
You may assume all of these values are positive integers less than 1001 and weight1 < weight2 < weight3 . The values represent the rate table below:
There will then be 1 or more lines of customer package sizes. Each of these will be a positive integer less than 1001.
The end of customer package sizes is indicated by the single integer 0.
The end of input will be indicated by end of file.
Output
For each input set, print the input set number.
Then, for each of the customer package sizes in the input set, create a line of output formatted as follows:
Weight ( pounds) is the number of pounds to be added to the package to obtain the price ( must be greater than or equal to 0). If more than one different weight results in the best possible price, use the smaller weight.
Where
Have a blank line after the output for each input set.
Sample Input
9 10
49 5
99 3
2
8
10
90
100
200
0
10 10
20 20
30 30
100
1
12
29
50
0
Sample Output
Set number 1: Weight (8) has best price $50 (add 2 pounds) Weight (10) has best price $50 (add 0 pounds) Weight (90) has best price $200 (add 10 pounds) Weight (100) has best price $200 (add 0 pounds) Weight (200) has best price $400 (add 0 pounds) Set number 2: Weight (1) has best price $10 (add 0 pounds) Weight (12) has best price $240 (add 0 pounds) Weight (29) has best price $870 (add 0 pounds) Weight (50) has best price $5000 (add 0 pounds)/*****************************************************************************************************************/
如上所示，标红加粗的全是重点
在本题中，除了基本的输入输出格式要注意外，个人以为还有另外两个问题需要格外关照
一是贪心法的选择；
利用贪心法解这道题目，只需考虑两种情况，一种是重物在当前区间内所需的花费大于重物在下一区间中所需的花费，则需选择下一区间；另一种则相反
二是“If more than one different weight results in the best possible price, use the smaller weight.”这条要求的注意
详见代码
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// Problem#: 1093
#include
#include
using namespace std;
#define INF 1001
struct Rate_Structure
{
int lower;
int upper;
int rate;
};
int main()
{
Rate_Structure ratestructure[4];
ratestructure[0].lower = 0;
ratestructure[3].upper = INF;
int counter = 1;
while(scanf("%d",&ratestructure[0].upper) != EOF)
{
for(int i = 0; i < 4; i++)
{
cin >> ratestructure[i].rate;
if(i < 3) //好神奇，他们两个竟然是错开的
{
ratestructure[i+1].lower = ratestructure[i].upper + 1;
}
if(i < 2)
{
cin >> ratestructure[i+1].upper;
}
}
int weight;
cout << "Set number " << counter << ":" << endl;
counter++;
while(cin >> weight && weight)
{
int cost, balance = 0;
for(int i = 0; i < 4; i++)
{
if(weight >= ratestructure[i].lower && weight <= ratestructure[i].upper)
{
cost = weight * ratestructure[i].rate;
}
else if(weight < ratestructure[i].lower)
{
int temp1 = cost;
cost = min(temp1, ratestructure[i].lower * ratestructure[i].rate);
if(cost == temp1)
{
balance = 0;
}
else
{
balance = ratestructure[i].lower - weight;
}
break;
}
}
cout << "Weight ("<< weight << ") has best price $" << cost << " (add " << balance << " pounds)" << endl;
}
cout << endl;
}
return 0;
}