Proud Merchants (关于背包顺序的一个反思)

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

背包问题状态从j转移到j-a[i]时j并未真正的出现是在后面的循环中实现的

所以背包是一个逆序的过程

所以进行反向贪心

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<iostream>
#include<set>
using namespace std;
struct node
{
    int q;
    int p;
    int v;
}a[1006];
bool cmp1(node x,node y)
{
    return x.q-x.p<y.q-y.p;
}
int dp[5005];
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(dp,0,sizeof(dp));
        for(int i=1;i<=n;i++)
            scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);
        sort(a+1,a+1+n,cmp1);
        cout<<endl;
        for(int i=1;i<=n;i++) cout<<a[i].p<<" "<<a[i].q<<endl;
        cout<<endl;
        for(int i=1;i<=n;i++)
        {
            for(int j=m;j>=0;j--)
            {
                if(j>=a[i].q) dp[j]=max(dp[j-a[i].p]+a[i].v,dp[j]);
            }
        }
        printf("%d
",dp[m]);
    }
    return 0;
}