hdu 3466 Proud Merchants（有排序的01背包）

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 4039    Accepted Submission(s): 1677

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   |   We have carefully selected several similar problems for you:  3460 3463 3468 3467 3465 

 

简单的01背包，排个序就好，代码很简单。

复制一下网上大神的思路：

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，那么就是p1+q2 > p2+q1，变形之后就是q1-p1 < q2-p2。

所以要针对每个属性的q-p来进行排序

题意：每个物品有p、q、v，三个属性，每个物品的话费为p，但是前提是必须有q，v则是得到的价值。

附上代码：

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
struct node
{
    int p,q,v;
} ss[505];

int maxs(int a,int b)
{
    return a>b?a:b;
}
bool cmp(node a,node b)
{
    return a.q-a.p<b.q-b.p;  //关键排序，详见上文分析
}
int main()
{
    int n,m,i,j;
    int dp[5005];
    while(~scanf("%d%d",&n,&m))
    {
        for(i=0; i<n; i++)
            scanf("%d%d%d",&ss[i].p,&ss[i].q,&ss[i].v);
        sort(ss,ss+n,cmp);
        memset(dp,0,sizeof(dp));
        for(i=0; i<n; i++)
            for(j=m; j>=ss[i].q; j--)
                dp[j]=maxs(dp[j],dp[j-ss[i].p]+ss[i].v);
        printf("%d
",dp[m]);
    }
    return 0;
}