HDU-3466-Proud Merchants(01背包问题)

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

要先按照Q-P排序，然后在按01背包处理

为什么要排序呢？我们应该先买Q-P最大的物体，然后买第二大、第三大、、、 最后买最小的。这个性质是这个问题的关键。
代码：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <string>
#include <cmath>
#include <algorithm>
#define N 504
#define INF 1000000001
#define ll long long
using namespace std;
struct node
{
    int price, vaule, q;
    friend bool operator < (node a, node b)
    {
        return a.q-a.price < b.q-b.price;
    }
}aa[N];
int dp[5005];
int main()
{
#ifndef ONLINE_JUDGE
    freopen("1.txt", "r", stdin);
#endif
    ios::sync_with_stdio(false);
    cin.tie(0);
    int i, j, k;
    int n, m;
    while(cin >> n >> m)
    {
        for (i = 0; i < n; i++)
            cin >> aa[i].price >> aa[i].q >> aa[i].vaule;
        sort(aa, aa+n);
        memset(dp, 0, sizeof(dp));
        for (i = 0; i < n; i++)
            for (j = m; j >= aa[i].q; j--)
                dp[j] = max(dp[j], dp[j-aa[i].price]+aa[i].vaule);
        cout << dp[m] << endl;
    }
    return 0;
}