Problem X

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

题意：类似的01背包问题，一开始看着不难，就写了但是案例就是不通过，看了好几天，才发现先原来购买顺序也会影响结果的；

解题思路：很显然，你会先判断q大，p小的物品买不买，对吧，因为在你价值最大，你应该尽可能的判断q大，p小的物品； 所以对q-p进行从小到大的排序，然后在进行DP；

感悟：今晚心情很差，只能写题；

代码：

#include

#include

#include

#include

#define maxn 5050

using namespace std;

struct node

{

    int
p,q,v;

    bool
operator < (const node & other) const

    {

       
return this->q-this->p

    }

};

int main()

{

   
//freopen("in.txt", "r", stdin);

    int
n,m;

   
while(~scanf("%d%d",&n,&m))

    {

       
int dp[maxn]={0};

       
node nod[maxn];

       
for(int i=0;i

           
scanf("%d%d%d",&nod[i].p,&nod[i].q,&nod[i].v);

       
sort(nod,nod+n);

       
for(int i=0;i

           
for(int j=m;j>=nod[i].p;j--)

       
{

           
if(j>=nod[i].q)

               
dp[j]=max(dp[j],dp[j-nod[i].p]+nod[i].v);

       
}

       
printf("%d\n",dp[m]);

    }

    return
0;

}