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最新推荐文章于 2019-11-24 23:57:32 发布

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最新推荐文章于 2019-11-24 23:57:32 发布

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分类专栏： ACM

本文链接：https://blog.youkuaiyun.com/xu_acm/article/details/51447099

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本文介绍了一种使用动态规划解决特定物品选择问题的方法。详细解释了输入格式、输出要求和示例输入输出，提供了相应的代码实现。通过分析每件商品的价格、价值和最低购买金额，目标是最大化在给定预算内获得的价值。

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题目：

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

题意：N件商品，钱共有M。。商品的价格为Pi，商品的价值为Vi，然后如果你想购买这件商品手中至少要有Qi元，让求最大能得到的价值。。

想法：动态规划。。方程是dp[j]=max(dp[j],dp[j-node[i].p]+node[i].v)

代码：

#include<stdio.h>
#include<string.h>
#include<algorithm>
#include<iostream>
using namespace std;
const int MAXN=550;
struct Node
{
int p,q,v;
}node[MAXN];
int inline max(int a,int b)
{
return a>b?a:b;
}
int dp[5500];
bool cmp(Node a,Node b)
{
return a.q-a.p < b.q-b.p;
}
int main()
{
int n,i,m;
//freopen("r.txt","r",stdin);
while(cin>>n>>m)
{
for(i=0;i<n;i++)
cin>>node[i].p>>node[i].q>>node[i].v;
sort(node,node+n,cmp);
memset(dp,0,sizeof(dp));
for(i=0;i<n;i++)
for(int j=m;j>=node[i].q;j--)
dp[j]=max(dp[j],dp[j-node[i].p]+node[i].v);
cout<<dp[m]<<endl;
}
return 0;
}

好难。。