hdu 3466 Proud Merchants 01背包变形

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 1978    Accepted Submission(s): 792

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

 

题意： 购买物品，pi qi vi  （pi代表要花的钱，qi代表你当前的钱要大于等于这个值才能买，vi 价值了。）

　　    和单纯的01背包比较，多了qi，在这里就发生区别了。

          for(j=sum_money;j>=f[i].qi&&(j-f[i].vi)>=0;j--)

　　　 采取的方式是 对 qi-pi进行从小到大排序，使得差值 越小的越先放。

         

　　　 为什么要这样排序呢？动手画一下。详细的证明，以后再写。

         

          其实单纯的01背包它的差值就是0，所以还是有共性的。

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;

struct node
{
    int pi,qi,vi;
}f[502];
int dp[5003];

bool cmp(node n1,node n2)
{
    return (n1.qi-n1.pi)<(n2.qi-n2.pi);
}

int main()
{
    int n,sum_money,max;
    int i,j,tmp;
    while(scanf("%d%d",&n,&sum_money)>0)
    {
        for(i=1;i<=n;i++)
        {
            scanf("%d%d%d",&f[i].pi,&f[i].qi,&f[i].vi);
        }
        sort(f+1,f+1+n,cmp);
        memset(dp,0,sizeof(dp));

        for(i=1;i<=n;i++)
        {
            for(j=sum_money;j>=f[i].qi&&(j-f[i].pi)>=0;j--)
            {
                tmp=dp[j-f[i].pi]+f[i].vi;
                if(tmp>dp[j])
                dp[j]=tmp;
            }
        }
        printf("%d\n",dp[sum_money]);
    }
    return 0;
}