F - Proud Merchants

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? 

InputThere 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. 

OutputFor 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

#include <iostream>
#include<stdio.h>
#include<algorithm>
#include<string.h>
using namespace std;
struct node
{
    int v1;
    int c;
    int w;
}a[100000];
int dp[10000];
bool cmp(node a,node b)
{
    return a.c-a.v1<b.c-b.v1;
}
int main()
{
     int n,v;
   while(~scanf("%d%d",&n,&v))
   {

memset(dp,0,sizeof dp);
     for(int i=0;i<n;i++)
     {
         scanf("%d%d%d",&a[i].v1,&a[i].c,&a[i].w);

     }
     sort(a,a+n,cmp);
     for(int i=0;i<n;i++)
        for(int j=v;j>=a[i].c;j--)
     {
         dp[j]=max(dp[j],dp[j-a[i].v1]+a[i].w);
     }
     printf("%d\n",dp[v]);
   }
    return 0;
}

排序的地方注意一下。。。因为A 需要的时间 A。V1+B.c     B的时间是B.V1+A.C;