HDU 3466 Proud Merchants（01背包+排序）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7457    Accepted Submission(s): 3117

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

 

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#include<cstdio>
#include<cstring>
#include<cmath>
#include<algorithm>
#include<queue>
#include<iostream>
using namespace std;
struct node
{
    int p,q,v;
    bool operator<(const node &a)const
    {
        return (a.q-a.p)>q-p;
    }
}a[550];
int dp[5050];
int main()
{
    int n,m,i,j;
    while(scanf("%d %d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
        {
            scanf("%d %d %d",&a[i].p,&a[i].q,&a[i].v);
        }
        sort(a,a+n);
        //for(i=0;i<n;i++)
            //printf("%d %d %d\n",a[i].p,a[i].q,a[i].v);
        for(i=0;i<n;i++)
            for(j=m;j>=a[i].q;j--)
            dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v);
        printf("%d\n",dp[m]);
    }
    return 0;
}