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HDU--3466--Proud Merchants--01背包

2024-08-05 14:24:59   222人阅读

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 2777    Accepted Submission(s): 1155


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
 
题意：给你N个物品和M单位的金钱，每个物品有价格a、购买的前提资金b、价值c，要买某个物品，你的钱必须大于等于b，求出最大能获得的价值量

这是个有前提的背包问题，主要是判断物品出场的顺序，本人表示跟上一题一样，准备作死的节奏，就不解析了，等弄明白了再加上



#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
#define Max(a,b) a>b?a:b
using namespace std;
struct node
{
    int a,b,c;
}s[555];
bool cmp(const node &a,const node &b)
{
    return a.b-a.a<b.b-b.a;	//排序规则
}
int main (void)
{
    int n,m,i,j,k,l;
    int dp[5555];
    while(scanf("%d%d",&n,&m)!=EOF)
    {
        for(i=0;i<n;i++)
        {
            scanf("%d%d%d",&s[i].a,&s[i].b,&s[i].c);
        }
        sort(s,s+n,cmp);
        memset(dp,0,sizeof(dp));
        for(i=0;i<n;i++)	//01背包
        for(j=m;j>=s[i].b;j--)
        dp[j]=Max(dp[j],dp[j-s[i].a]+s[i].c);
        printf("%d\n",dp[m]);
    }
    return 0;
}

总结：今天就做了这两个题，一个是K大值的背包，一个是有前提的背包，这几天有得溜了！


HDU--3466--Proud Merchants--01背包


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