HDU3466 Proud Merchants 01背包通过排序去除后效性

最新推荐文章于 2020-09-14 22:11:51 发布

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#01背包消除后效性

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本文介绍了一种解决特定背包问题的方法，通过合理排序商品并使用动态规划算法，以求得在有限资金下能获得的最大价值。适用于算法竞赛及动态规划初学者。

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 8234 Accepted Submission(s): 3457

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

因为每个物品都有一个限制q，而01背包dp是从前i个物品转移到前i+1个物品，所以必须保证转移时无后效性，即前面i个物品选择的结果不会影响到后面的选择，在这里体现在前i个物品中会影响的范围为q-p;如果第i+1个物品的q-p > 前i个物品的q-p,则可以保证如果选择第i个物品，那么前面i+1个物品的q的限制条件都能满足。 即 qi+1 在的位置>qi

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;
const int maxn = 510;
const int INF = 0x3f3f3f3f;

int n,m;
int dp[5005];
struct Node
{
  int p,q,v;
}a[maxn];

bool cmp (Node a, Node b)
{
  return a.q - a.p < b.q - b.p;
}

int main()
{
  while(~scanf("%d%d",&n,&m))
  {
    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);
    memset(dp,0,sizeof(dp));
    for(int i = 0; i < n; i++)
       for(int j = m; j >= a[i].q; j--)
            dp[j] = max(dp[j],dp[j-a[i].p] + a[i].v);
    printf("%d\n",dp[m]);
  }
  return 0;
}