【杭电oj】3466 - Proud Merchants（01背包，排序处理）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4481    Accepted Submission(s): 1859

Problem Description

Recently, iSea went to an ancient country. For such a long time, it was the most wealthy and powerful kingdom in the world. As a result, the people in this country are still very proud even if their nation hasn’t been so wealthy any more.
The merchants were the most typical, each of them only sold exactly one item, the price was Pi, but they would refuse to make a trade with you if your money were less than Qi, and iSea evaluated every item a value Vi.
If he had M units of money, what’s the maximum value iSea could get?

 

Input

There are several test cases in the input.

Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.
Then N lines follow, each line contains three numbers Pi, Qi and Vi (1 ≤ Pi ≤ Qi ≤ 100, 1 ≤ Vi ≤ 1000), their meaning is in the description.

The input terminates by end of file marker.

 

Output

For each test case, output one integer, indicating maximum value iSea could get.

 

Sample Input

   
   
   
   

    
    
    
    2 10
10 15 10
5 10 5
3 10
5 10 5
3 5 6
2 7 3
   
   
   
   

 

Sample Output

   
   
   
   

    
    
    
    5
11
   
   
   
   

 

Author

iSea @ WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU

这道题的精髓就在于对于物品的排序上了，如果直接用dp的话，有些物品需要的钱数太高，如果靠后放入背包的话，则不能取走，这样就影响了最优解。

p - q 处理的意思应该为：把价值较高的物品靠前考虑是否放入背包。

代码如下：

#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
	int p,q,v;		//花费，限制条件（钱不够不让买），价值 
}c[555];
bool cmp (node a,node b)		//差额越大越优先（价值大） 
{
	return a.p - a.q > b.p - b.q;		//这个排序方法很巧，但是不是太懂原理 
}
int dp[5555];
int main()
{
	int n,m;		//物品数、钱数
	while (~scanf ("%d %d",&n,&m))
	{
		memset (dp,0,sizeof (dp));
		for (int i = 1 ; i <= n ; i++)
			scanf ("%d %d %d",&c[i].p,&c[i].q,&c[i].v);
		sort (c+1,c+1+n,cmp);
		for (int i = 1 ; i <= n ; i++)
		{
			for (int j = m ; j >= c[i].q ; j--)
			{
				dp[j] = max (dp[j] , dp[j-c[i].p] + c[i].v);
			}
		}
		printf ("%d\n",dp[m]);
	}
	return 0;
}