Proud Merchants（POJ 3466 01背包+排序）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4257    Accepted Submission(s): 1757

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

题目大意：n中商品，m元钱，每种商品都有p,q,v属性，p价格，q表示买这种商品你需要带q元老板才愿意和你交易，v这种商品的实际价值。求问最多可以获得多少价值

思路：此题对于第二个样例，第一件商品 5 10 5只会把dp[10]更新出来，但实际上花费了5，更新第二个商品时，需要dp[10]=max(dp[10-5]+6,dp[10])，此时需要借助上一层的dp[5],但实际上此时dp[5]还没有更新。所以实际上对于一个p,q他最小能跟新出dp[q-p]，所以需要对每件商品安装q-p大小排序，然后再背包求解

#include <cstring>
#include <algorithm>
#include <cstdio>
#include <iostream>
using namespace std;
int n,m;
struct node
{
    int p,q,v;
}a[501];
int dp[5002];
bool cmp(node a,node b)
{
    return a.q-a.p<=b.q-b.p;
}
int main()
{
    int i,j;
    freopen("in.txt","r",stdin);
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)
            scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);
        sort(a,a+n,cmp);
        for(i=0;i<n;i++)
        {
            for(j=m;j>=a[i].q;j--)
                dp[j]=max(dp[j],dp[j-a[i].p]+a[i].v);
        }
        printf("%d
",dp[m]);
    }
}