HDOJ 3466 Proud Merchants 【0 1揹包】

Proud Merchants

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)

Total Submission(s): 3859    Accepted Submission(s): 1615

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

題目鏈接：HDOJ 3466 Proud Merchants 【0 1揹包】

題意：首先給出物品數量和手中資金人後每樣物品給出價格，需要購買時手中至少需要多少資金，還有物品本身的價值要求求出最大資金

思路：0 1 揹包，要注意的是要先按 q-p 排序（使差值最小）

已AC代碼：

#include<cstdio>
#include<cstring>
#include<algorithm>
#define Max(x,y) (x>y?x:y)
#define M 6000
using namespace std;

struct node{
	int p,q,v;
}s[M];
int dp[M],n,m;

bool cmp(node a,node b)
{
	return a.q-a.p < b.q-b.p;
}
int main()
{
	int i,j;
	while(scanf("%d%d",&n,&m)!=EOF)
	{
		for(i=0;i<n;++i)
			scanf("%d%d%d",&s[i].p,&s[i].q,&s[i].v);
			
		sort(s,s+n,cmp);
		
		memset(dp,0,sizeof(dp));
		
		for(i=0;i<n;++i)
			for(j=m;j>=s[i].q;--j)
				dp[j]=Max(dp[j],dp[j-s[i].p]+s[i].v);
				
		printf("%d\n",dp[m]);
	}
	return 0;
}