Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 6633    Accepted Submission(s): 2769

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   Author iSea @ WHU     Source 2010 ACM-ICPC Multi-University Training Contest（3）——Host by WHU   Recommend zhouzeyong   |   We have carefully selected several similar problems for you:  3460 3463 3468 3467 3465      题解来自sdfzchy，为什么不自己写题解呢？之前屯题一直没有更博客，等到想更的时候才发现……真烦…… 题目大意：n个商品，m元钱，每种商品都有p,q,v，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
题解： 01背包的变形；
假设有2个物品，A（P1,Q1,V1），B(P2,Q2,V2)。
如果先买A后买B需要（不等于花费）P1+Q2，如果先买B后买A需要P2+Q1，选择两者之中较小的能保证遍历所有情况。
若先买A，则P1+Q2 < P2+Q1 ,即Q2-P2 < Q1-P1。故应先买q-p较大的
考虑0/1背包中的更新其实是逆向的，所以要把q-p从小到大排序，再做01背包。  

 1 #include
 2 #include
 3 #include
 4 #include
 5 using namespace std;
 6 int n,m;
 7 int f[50010];
 8 struct dt{
 9     int p,q,v;
10     bool operator<(const dt&aa)const{
11         return q-paa.p;
12     }
13 }data[510];
14 int main(){
15     while(scanf("%d%d",&n,&m)!=EOF){
16         memset(f,0,sizeof(f)); 
17         for(int i=1;i<=n;i++) scanf("%d%d%d",&data[i].p,&data[i].q,&data[i].v);
18         sort(data+1,data+n+1);
19         for(int i=1;i<=n;i++)
20             for(int j=m;j>=max(data[i].q,data[i].p);j--)
21                 f[j]=max(f[j],f[j-data[i].p]+data[i].v);
22         printf("%d\n",f[m]);        
23     }
24     return 0;
25 }