ACM HDU 3466 Proud Merchants（某种顺序下的背包）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 925    Accepted Submission(s): 368

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

 

Recommend

zhouzeyong

 

这是一道01背包的变形题，题目增加了一个限制条件，即当你所拥有的钱数大于某个限定值时才可以购买该物品。

按照q - p以由大到小的顺序排序，然后进行01背包的DP即可。

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=5005;
int dp[MAXN];
struct Node
{
    int p,q,v;
}node[505];
bool cmp(Node a,Node b)
{
    return  (a.q-a.p)<(b.q-b.p);
}    
int main()
{
    int n,m;
    int i,j;
    int p,q,v;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<=m;i++)
          dp[i]=0;
        for(i=0;i<n;i++)
        {
            scanf("%d%d%d",&node[i].p,&node[i].q,&node[i].v);
        }    
        sort(node,node+n,cmp);
        for(i=0;i<n;i++)
        {
            for(j=m;j>=node[i].p;j--)
            {
                if(j>=node[i].q)
                  dp[j]=max(dp[j],dp[j-node[i].p]+node[i].v);
            }    
        }
        int ans=0;
        for(i=1;i<=m;i++)
          if(ans<dp[i])  ans=dp[i];
        printf("%d\n",ans);
              
    }    
    return 0;
}