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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.<br>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.<br>If he had M units of money, what’s the maximum value iSea could get?<br><br>

 

Input

There are several test cases in the input.<br><br>Each test case begin with two integers N, M (1 ≤ N ≤ 500, 1 ≤ M ≤ 5000), indicating the items’ number and the initial money.<br>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.<br><br>The input terminates by end of file marker.<br><br>

 

Output

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

 

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

 

大意：n中商品，m元钱，每种商品都有p,q,v属性，p价格，q表示买这种商品你需要带q元老板才愿意和你交易，v这种商品的实际价值。求问最多可以获得多少价值

思路：此题对于第二个样例，第一件商品 5 10 5只会把dp[10]更新出来，但实际上花费了5，更新第二个商品时，需要dp[10]=max(dp[10-5]+6,dp[10])，此时需要借助上一层的dp[5],但实际上此时dp[5]还没有更新。所以实际上对于一个p,q他最小能跟新出dp[q-p]，所以需要对每件商品安装q-p大小排序，然后再背包求解

代码：

#include<iostream>
#include<algorithm>
#include<cstdio>
#include<cstring>
using namespace std;
struct node
{
    int p,q,v;
}iarr[550];
int f[550][5500],dp[5003];
int cmp(const node &a, const node &b)
{
    return a.q-a.p<b.q-b.p;
}
int main()
{
    int n,m;
    while(cin>>n>>m)
    {
        for(int i=1;i<=n;++i)
           cin>>iarr[i].p>>iarr[i].q>>iarr[i].v;
        sort(iarr+1,iarr+n+1,cmp);
        memset(dp,0,sizeof(dp));
        memset(f,0,sizeof(f));
        for(int i=1;i<=n;++i)
            for(int j=0;j<=m;++j)
            {
                f[i][j]=f[i-1][j];


        if(j>=iarr[i].q) f[i][j]=max(f[i][j],f[i-1][j-iarr[i].p]+iarr[i].v);
            }
        cout<<f[n][m]<<endl;


    }
    return 0;
}