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
【分析】： 需要对物品按 qi-pi 的值从小到大排序，因为这样可以保证每次更新的状态值从小到大递增，前面更新过的状态不会影响后面更新的状态。
每件物品有一个限制，只有当你当前现金大于qi的时候才会卖给你。
这题好好想了一下，跟之前做过的一道题有些类似。考虑简化版，有两个物品(p1,q1,v1),(p2,q2,v2)，然后物品1先放的话，物品2就可以借助物品1产生的各种状态来进行下一步转移，而如果物品2的q2值过高，在这个[q2,m]的区间内都不存在物品1造成的新状态的话，那么物品1的状态就没有得到利用。而如果交换顺序，先放了物品2，那么显然物品1就可以利用物品2产生的新状态。
所以物品1能从物品2转移的状态区间其实是[min(q1+p2,m),m]，物品2能从物品1转移的状态区间是[min(q2+p1,m),m]。所以尽可能地复用这个区间，让区间小的先来，区间大的后来，这样排序之后所有物品都能从前面的物品得到新状态进行转移。
而普通的01背包之所以不需要排序，是因为p1q1,p2q2，排序跟不排是一回事。这一类的dp题要注意后效性是否存在，如果存在通过改变顺序之类的办法来取消后效性。再有杭电上饭卡那题，qi恒定为5，所以也是需要排序的。
#include<cstdio>
#include<string>
#include<cstdlib>
#include<cmath>
#include<iostream>
#include<cstring>
#include<set>
#include<queue>
#include<algorithm>
#include<vector>
#include<map>
#include<cctype>
#include<stack>
#include<sstream>
#include<list>
#include<assert.h>
#include<bitset>
#include<numeric>
#define debug() puts("++++")
#define gcd(a,b) __gcd(a,b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define fi first
#define se second
#define pb push_back
#define sqr(x) ((x)*(x))
#define ms(a,b) memset(a,b,sizeof(a))
#define sz size()
#define be begin()
#define pu push_up
#define pd push_down
#define cl clear()
#define lowbit(x) -x&x
#define all 1,n,1
#define rep(i,n,x) for(int i=(x); i<(n); i++)
#define in freopen("in.in","r",stdin)
#define out freopen("out.out","w",stdout)
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int,int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e18;
const int maxn = 1e5 + 5;
const int maxm = 1e6 + 10;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int dx[] = {-1,1,0,0,1,1,-1,-1};
const int dy[] = {0,0,1,-1,1,-1,1,-1};
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
int n,m;
#define S 100000
#define M 200000
int dp[maxn];
struct node
{
int p,q,v;
int sub;
}a[maxn];
bool cmp(node x, node y)
{
return x.sub < y.sub;
}
int main()
{
while(cin >> n >> m)
{
ms(dp,0);
for(int i=1;i<=n;i++) //但是如果你的钱少于Qi，他们会拒绝进行交易
{
cin >> a[i].p >> a[i].q >> a[i].v;
a[i].sub = a[i].q - a[i].p;
}
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] << endl;
}
return 0;
}
/*
3 10
3 5 6
2 7 3
5 10 5
*/