HDU--3466--Proud Merchants--01背包

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others) Total Submission(s): 2777 Accepted Submission(s): 1155

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

題意：給你N個物品和M單位的金錢，每個物品有價格a、購買的前提資金b、價值c，要買某個物品，你的錢必須大於等於b，求出最大能獲得的價值量

這是個有前提的背包問題，主要是判斷物品出場的順序，本人表示跟上一題一樣，准備作死的節奏，就不解析了，等弄明白了再加上

#include 
#include 
#include 
#include 
#define Max(a,b) a>b?a:b
using namespace std;
struct node
{
    int a,b,c;
}s[555];
bool cmp(const node &a,const node &b)
{
    return a.b-a.a=s[i].b;j--)
        dp[j]=Max(dp[j],dp[j-s[i].a]+s[i].c);
        printf("%d\n",dp[m]);
    }
    return 0;
}

總結：今天就做了這兩個題，一個是K大值的背包，一個是有前提的背包，這幾天有得溜了！