poj1260 dp
2015-03-19 14:22
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如题:http://poj.org/problem?id=1260
Pearls
Description
In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name because it delivers to the royal family of Pearlania. But it also produces
bracelets and necklaces for ordinary people. Of course the quality of the pearls for these people is much lower then the quality of pearls for the royal family.In Pearlania pearls are separated into 100 different quality classes. A quality class is identified
by the price for one single pearl in that quality class. This price is unique for that quality class and the price is always higher then the price for a pearl in a lower quality class.
Every month the stock manager of The Royal Pearl prepares a list with the number of pearls needed in each quality class. The pearls are bought on the local pearl market. Each quality class has its own price per pearl, but for every complete deal in a certain
quality class one has to pay an extra amount of money equal to ten pearls in that class. This is to prevent tourists from buying just one pearl.
Also The Royal Pearl is suffering from the slow-down of the global economy. Therefore the company needs to be more efficient. The CFO (chief financial officer) has discovered that he can sometimes save money by buying pearls in a higher quality class than is
actually needed.No customer will blame The Royal Pearl for putting better pearls in the bracelets, as long as the
prices remain the same.
For example 5 pearls are needed in the 10 Euro category and 100 pearls are needed in the 20 Euro category. That will normally cost: (5+10)*10+(100+10)*20 = 2350 Euro.Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.
The problem is that it requires a lot of computing work before the CFO knows how many pearls can best be bought in a higher quality class. You are asked to help The Royal Pearl with a computer program.
Given a list with the number of pearls and the price per pearl in different quality classes, give the lowest possible price needed to buy everything on the list. Pearls can be bought in the requested,or in a higher quality class, but not in a lower one.
Input
The first line of the input contains the number of test cases. Each test case starts with a line containing the number of categories c (1<=c<=100). Then, c lines follow, each with two numbers ai and pi. The first of these numbers
is the number of pearls ai needed in a class (1 <= ai <= 1000).
The second number is the price per pearl pi in that class (1 <= pi <= 1000). The qualities of the classes (and so the prices) are given in ascending order. All numbers in the input are integers.
Output
For each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.
Sample Input
Sample Output
Source
Northwestern Europe 2002
题目大意:买珍珠,给出每一种珍珠的需要的数量和价钱,如果买这种珍珠,必须多买10个,价格低的珍珠可以用高的顶上去。问买这些珍珠的最小花费。
思路:dp[i]:买前i种珍珠的最小花费。
我一开始的dp思路是 dp[i]=min(dp[i-1]+(a[i]+10)*p[i],dp[i]).
dp[i]初始化为(前i件物品的总数量+10)*10.
也就是前i中的最佳策略=前i-1种的最佳策略+单独买第i种,用第i种买所有前i种珍珠。题目给出价值升序。
之后发现这种方法不对,原因在不能仅仅省了这一次的钱却可能使得之后的最小方案被舍弃。
因此修改策略。
dp[i]=min(dp[j]+(sum[i]-sum[j]+10)*p[i],dp[i]). j∈[1,n)
双重循环保证后效性。
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define MAXN 105
#define min(a,b)(a<b?a:b)
int dp[MAXN]; //dp[i]:买前i件物品的最小花费。
int n;
int a[MAXN];
int p[MAXN];
int sum[MAXN];
int main()
{
// freopen("C:\\1.txt","r",stdin);
int T;
cin>>T;
while(T--)
{
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
memset(p,0,sizeof(p));
memset(sum,0,sizeof(sum));
cin>>n;
int i,j;
for(i=1;i<=n;i++)
{
scanf("%d%d",&a[i],&p[i]);
sum[i]=sum[i-1]+a[i];
}
for(i=1;i<=n;i++)
{
dp[i]=(sum[i]+10)*p[i];
for(j=1;j<i;j++)
dp[i]=min(dp[i],dp[j]+(sum[i]-sum[j]+10)*p[i]);
}
printf("%d\n",dp
);
}
return 0;
}
Pearls
Time Limit: 1000MS | Memory Limit: 10000K | |
Total Submissions: 7698 | Accepted: 3806 |
In Pearlania everybody is fond of pearls. One company, called The Royal Pearl, produces a lot of jewelry with pearls in it. The Royal Pearl has its name because it delivers to the royal family of Pearlania. But it also produces
bracelets and necklaces for ordinary people. Of course the quality of the pearls for these people is much lower then the quality of pearls for the royal family.In Pearlania pearls are separated into 100 different quality classes. A quality class is identified
by the price for one single pearl in that quality class. This price is unique for that quality class and the price is always higher then the price for a pearl in a lower quality class.
Every month the stock manager of The Royal Pearl prepares a list with the number of pearls needed in each quality class. The pearls are bought on the local pearl market. Each quality class has its own price per pearl, but for every complete deal in a certain
quality class one has to pay an extra amount of money equal to ten pearls in that class. This is to prevent tourists from buying just one pearl.
Also The Royal Pearl is suffering from the slow-down of the global economy. Therefore the company needs to be more efficient. The CFO (chief financial officer) has discovered that he can sometimes save money by buying pearls in a higher quality class than is
actually needed.No customer will blame The Royal Pearl for putting better pearls in the bracelets, as long as the
prices remain the same.
For example 5 pearls are needed in the 10 Euro category and 100 pearls are needed in the 20 Euro category. That will normally cost: (5+10)*10+(100+10)*20 = 2350 Euro.Buying all 105 pearls in the 20 Euro category only costs: (5+100+10)*20 = 2300 Euro.
The problem is that it requires a lot of computing work before the CFO knows how many pearls can best be bought in a higher quality class. You are asked to help The Royal Pearl with a computer program.
Given a list with the number of pearls and the price per pearl in different quality classes, give the lowest possible price needed to buy everything on the list. Pearls can be bought in the requested,or in a higher quality class, but not in a lower one.
Input
The first line of the input contains the number of test cases. Each test case starts with a line containing the number of categories c (1<=c<=100). Then, c lines follow, each with two numbers ai and pi. The first of these numbers
is the number of pearls ai needed in a class (1 <= ai <= 1000).
The second number is the price per pearl pi in that class (1 <= pi <= 1000). The qualities of the classes (and so the prices) are given in ascending order. All numbers in the input are integers.
Output
For each test case a single line containing a single number: the lowest possible price needed to buy everything on the list.
Sample Input
2 2 100 1 100 2 3 1 10 1 11 100 12
Sample Output
330 1344
Source
Northwestern Europe 2002
题目大意:买珍珠,给出每一种珍珠的需要的数量和价钱,如果买这种珍珠,必须多买10个,价格低的珍珠可以用高的顶上去。问买这些珍珠的最小花费。
思路:dp[i]:买前i种珍珠的最小花费。
我一开始的dp思路是 dp[i]=min(dp[i-1]+(a[i]+10)*p[i],dp[i]).
dp[i]初始化为(前i件物品的总数量+10)*10.
也就是前i中的最佳策略=前i-1种的最佳策略+单独买第i种,用第i种买所有前i种珍珠。题目给出价值升序。
之后发现这种方法不对,原因在不能仅仅省了这一次的钱却可能使得之后的最小方案被舍弃。
因此修改策略。
dp[i]=min(dp[j]+(sum[i]-sum[j]+10)*p[i],dp[i]). j∈[1,n)
双重循环保证后效性。
#include<iostream>
#include<cstring>
#include<cstdio>
using namespace std;
#define MAXN 105
#define min(a,b)(a<b?a:b)
int dp[MAXN]; //dp[i]:买前i件物品的最小花费。
int n;
int a[MAXN];
int p[MAXN];
int sum[MAXN];
int main()
{
// freopen("C:\\1.txt","r",stdin);
int T;
cin>>T;
while(T--)
{
memset(dp,0,sizeof(dp));
memset(a,0,sizeof(a));
memset(p,0,sizeof(p));
memset(sum,0,sizeof(sum));
cin>>n;
int i,j;
for(i=1;i<=n;i++)
{
scanf("%d%d",&a[i],&p[i]);
sum[i]=sum[i-1]+a[i];
}
for(i=1;i<=n;i++)
{
dp[i]=(sum[i]+10)*p[i];
for(j=1;j<i;j++)
dp[i]=min(dp[i],dp[j]+(sum[i]-sum[j]+10)*p[i]);
}
printf("%d\n",dp
);
}
return 0;
}
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