这道题目看出背包很容易，主要是处理背包的时候需要按照q-p排序然后进行背包。

这样保证了尽量多的利用空间。

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

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

#include <algorithm>
#include <iostream>
#include <stdlib.h>
#include <string.h>
#include <iomanip>
#include <stdio.h>
#include <string>
#include <queue>
#include <cmath>
#include <stack>
#include <map>
#include <set>
#define eps 1e-10
///#define M 1000100
#define LL __int64
///#define LL long long
///#define INF 0x7ffffff
#define INF 0x3f3f3f3f
#define PI 3.1415926535898
#define zero(x) ((fabs(x)<eps)?0:x)

///#define mod 10007

const int maxn = 5010;
using namespace std;

int dp[maxn];

struct node
{
    int p, q, v;
}f[510];

bool cmp(node a, node b)
{
    return a.q-a.p < b.q-b.p;
}

int main()
{
    int n, m;
    while(~scanf("%d %d",&n, &m))
    {
        for(int i = 1; i <= n; i++) scanf("%d %d %d",&f[i].p, &f[i].q, &f[i].v);
        for(int i = 0; i <= m; i++) dp[i] = 0;
        sort(f+1, f+n+1, cmp);
        for(int i = 1; i <= n; i++)
            for(int j = m; j >= f[i].q; j--) dp[j] = max(dp[j] , dp[j-f[i].p] + f[i].v);
        printf("%d\n",dp[m]);
    }
}

/*
3 10
3 6 10
3 8 4
2 10 7
*/