hdu3466 Proud Merchants（排序+01背包）

最新推荐文章于 2022-09-09 22:07:37 发布

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本文探讨了ProudMerchants问题，这是一个经典的计算机科学问题，涉及到贪心算法的应用。文章通过具体实例展示了如何利用动态规划来求解最大价值问题，适用于算法初学者及竞赛选手。

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Proud Merchants

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2145 Accepted Submission(s): 850

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

Sample Output

5
11

这个排序曾在贪心策略那里遇到过

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
struct ss
{
    int p,q,v;
} s[504];
int f[5004];
bool cmp(ss a,ss b)
{
    return a.q-a.p<b.q-b.p;
}
int main()
{
    int n,m,i,j;
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        memset(f,0,sizeof(f));
        for(i=0; i<n; i++)
            scanf("%d%d%d",&s[i].p,&s[i].q,&s[i].v);
        sort(s,s+n,cmp);
        for(i=0; i<n; i++)
            for(j=m; j>=s[i].q; j--)
            {
                int temp=f[j-s[i].p]+s[i].v;
                if(temp>f[j])
                    f[j]=temp;
            }
        printf("%d\n",f[m]);
    }
}