01背包 hdu3446

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5983    Accepted Submission(s): 2502

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

这道题可能难在为何要按照p-q排序。简单一项，其实就是 比如你有20，有三个东西 p,q分别为（5，10），（3，10），（3，9）

我们肯定先去买q大的，如果q大p还很小我们肯定要先买这个，花费的很小，美哉美哉。

所以我们就应该尽可能去先买q-p比较大的物品。

之后就变成了一个01背包，判断条件改成q的就行了。

因为dp师你想的，所以我们把q-p小的排在前面，大的排在后面，这样子结构到最底层才取的q-p最大的那个。

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

struct thing
{
    int p,q,v;
};
thing k[505];
int dp[5005];

bool cmp(thing a,thing b)
{
    if(a.q-a.p<b.q-b.p)
        return 1;
    else if(a.q-a.p==b.q-b.p)
        return a.v>b.v;
    else return 0;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=0;i<n;i++)
            cin>>k[i].p>>k[i].q>>k[i].v;
        sort(k,k+n,cmp);
        memset(dp,0,sizeof(dp));
        dp[0]=0;
        for(int i=0;i<n;i++)
            for(int j=m;j>=k[i].q;j--)
                dp[j]=max(dp[j],dp[j-k[i].p]+k[i].v);
        cout<<dp[m]<<endl;
    }
    return 0;
}