Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5566    Accepted Submission(s): 2345

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：

买一件价格为p的物品时手里的钱不少于p时才能购买。给出n件物品的p,q,v，v代表价值，总钱数m，问最大能够买到多大的价值。

代码：

//先买q大而p小的比较好，所以按照q-p排序，注意是从小到大排序，因为背包公式中
//f[i]=max(f[i],f[i-m]+v)是假设i-m被装满时再装下一件，是假设倒着装的。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[5005];
int n,m;
struct proud
{
    int p,q,v,ch;
};
bool cmp(proud a,proud b)
{
    return a.ch<b.ch;
}
/*void zeroonepack(int v,int m,int ttl)
{
    for(int i=ttl;i>=v;i--)
    f[i]=max(f[i],f[i-v]+m);
}
void complitpack(int v,int m,int ttl)
{
    for(int i=v;i<=ttl;i++)
    f[i]=max(f[i],f[i-v]+m);
}
void multipack(int v,int m,int c,int ttl)
{
    if(c*v>=ttl)
    complitpack(v,m,ttl);
    else
    {
        int k=1;
        while(k<c)
        {
            zeroonepack(k*v,k*m,ttl);
            c-=k;
            k*=2;
        }
        zeroonepack(c*v,c*m,ttl);
    }
}*/
int main()
{
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(f,0,sizeof(f));
        proud pr[505];
        for(int i=1;i<=n;i++)
        {
            scanf("%d%d%d",&pr[i].p,&pr[i].q,&pr[i].v);
            pr[i].ch=pr[i].q-pr[i].p;
        }
        sort(pr+1,pr+n+1,cmp);
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=pr[i].q;j--)
            {
                f[j]=max(f[j],f[j-pr[i].p]+pr[i].v);
            }
        }
        int ans=0;
        printf("%d\n",f[m]);
    }
    return 0;
}