hdu3466-Proud Merchants

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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

题意：买东西，每个东西有三个特征值，p代表价格，q代表你手中钱必须不低于q才能买这个物品，v代表得到的价值。
dp问题
这题因为涉及到q，所以不能直接就01背包了。因为如果一个物品是5 9，一个物品是5 6，对第一个进行背包的时候只有dp[9],dp[10],…,dp[m]，再对第二个进行背包的时候，如果是普通的，应该会借用前面的dp[8],dp[7]之类的，但是现在这些值都是0，所以会导致结果出错。于是要想到只有后面要用的值前面都可以得到，那么才不会出错。设A:p1,q1 B:p2,q2，如果先A后B，则至少需要p1+q2的容量，如果先B后A，至少需要p2+q1的容量，如果A排在B前面，那么就是p1+q2 > p2+q1，变形之后就是q1-p1 < q2-p2。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
const int M = 510;
struct node
{
    int p, q, v, m;
    friend bool operator < (node a, node b)
    {
        return a.m < b.m;
    }
}good[M];
int n, m;
int dp[M*10];
int main()
{
    int i, j, ans;
    while(~scanf("%d%d", &n, &m))
    {
        for (i = 0; i < n; i++)
        {
            scanf("%d%d%d", &good[i].p, &good[i].q, &good[i].v);
            good[i].m = good[i].q - good[i].p;
            if (good[i].q > m)
            {
                i--;
                n--;
            }
        }
        ans = 0;
        memset(dp, 0, sizeof(dp));
        sort(good, good+n);
        for (i = 0; i < n; i++)
        {
            for (j = m; j >= good[i].q; j--)
            {
                dp[j] = max(dp[j], dp[j-good[i].p]+good[i].v);
            }
        }
        cout << dp[m] << endl;
    }
    return 0;
}