0/1背包_1

最新推荐文章于 2022-03-21 23:06:56 发布

原创最新推荐文章于 2022-03-21 23:06:56 发布 · 283 阅读

0 ·

CC 4.0 BY-SA版权

算法 acm 专栏收录该内容

46 篇文章

订阅专栏

本文介绍了一种特殊的01背包问题，其特点在于交易条件的限制。通过对物品按差值（Qi-Pi）排序，利用01背包算法求解商人与iSea之间的最优交易方案，以获得最大价值。

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

这道题和一般的01背包不一样，因为一般的01背包与装物品的顺序无关，这道题所以要按照差值（qi-pi）从小到大排序，然后用01背包的方法进行操作就行了！

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<algorithm>
using namespace std;
struct Node
{
    int p,q,v;
}a[550];

int bag[5010];

bool cmp(Node a,Node b)
{
    return (a.q-a.p)<(b.q-b.p);
}
int main()
{
    int n,m;
    while(~scanf("%d%d",&n,&m))
    {
        memset(bag,0,sizeof(bag));
        for(int i=0;i<n;i++)
            scanf("%d%d%d",&a[i].p,&a[i].q,&a[i].v);
        sort(a,a+n,cmp);
        for(int i=0;i<n;i++)
        {
            for(int j=m;j>=a[i].q;j--)//等号是临界情况
            {
                bag[j]=max(bag[j],bag[j-a[i].p]+a[i].v);
            }
        }
            printf("%d\n",bag[m]);
    }
    return 0;
}