Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 5599    Accepted Submission(s): 2362

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

f[i][j]是前i个物品j元，只有j>qi时才能买第i个物品

所以考虑第i个物品，普通情况下最小可以从f[0]更新来，现在是f[qi-pi]，也就是说qi-pi被浪费了

这个浪费当然越小越好

A：p1,q1   B: p2,q2，先选A，则至少需要p1+q2的容量，而先选B则至少需要p2+q1，如果p1+q2>p2+q1，那么要选两个的话的就要先选A再选B，公式可换成q1-p1 < q2-p2，

首先发现qi-pi小，不可能有比qi-pi更小的qj，使得可以从j到i

这样可以保证后来的j-pi大于前面的q-p，也就是更新来的范围递增

就按照取出两个比较的套路来行了

//
//  main.cpp
//  hdu3466
//
//  Created by Candy on 29/10/2016.
//  Copyright © 2016 Candy. All rights reserved.
//

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <queue>
using namespace std;
const int N=505,M=5005;
inline int read(){
    char c=getchar();int x=0,f=1;
    while(c<‘0‘||c>‘9‘){if(c==‘-‘)f=-1; c=getchar();}
    while(c>=‘0‘&&c<=‘9‘){x=x*10+c-‘0‘; c=getchar();}
    return x*f;
}
int n,m;
struct item{
    int p,q,w;
    bool operator <(const item &r)const{return q-p<r.q-r.p;}
}a[N];
int f[M];
void dp(){
    memset(f,0,sizeof(f));
    for(int i=1;i<=n;i++)
        for(int j=m;j>=a[i].q;j--)
            f[j]=max(f[j],f[j-a[i].p]+a[i].w);
}
int main(int argc, const char * argv[]){
    while(scanf("%d%d",&n,&m)!=EOF){
        for(int i=1;i<=n;i++){a[i].p=read();a[i].q=read();a[i].w=read();}
        sort(a+1,a+1+n);
        dp();
        printf("%d\n",f[m]);
    }
    return 0;
}