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Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 7536    Accepted Submission(s): 3144

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

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　　分析：一开始考虑直接01背包，将限制条件直接换成j>=q[i]，但是发现这种思路是错的。显然物品的顺序会对其产生影响，那么就要考虑如何排序。然后试了几种排序发现都不对，还是K_lord和five20大佬讲过以后才明白。

　　设两个物品a,b，那么如果在选了一个物品以后，要让剩余的钱能买到的物品尽可能的多，则要满足qa+pb<qb+pa。或者这么说，假如你有m的金钱，而且m>qb>qa,m-pa>qb,m-pb>qa，那么如果先选b就不能选a，但如果先选a就可以选b，那么上面的式子成立。再变换一下得到qa-pa<qb-pb，那么就按照这个式子排序然后01背包就OK了。

　　Code：

#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cmath>
#include<algorithm>
#include<iostream>
#include<iomanip>
using namespace std;
const int N=5e3+7;
int n,m,dp[N];
struct Node{int p,q,v;}a[N];
bool cmp(Node x,Node y)
{return x.q-x.p<y.q-y.p;}
int main()
{
  ios::sync_with_stdio(false);
  while(cin>>n>>m){
    for(int i=1;i<=n;i++)
      cin>>a[i].p>>a[i].q>>a[i].v;
    memset(dp,0,sizeof(dp));
    sort(a+1,a+n+1,cmp);
    for(int i=1;i<=n;i++)
      for(int j=m;j>=a[i].q;j--)
    dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v);
    cout<<dp[m]<<"
";}
  return 0;
}