【杭电3466】贪心+背包

Proud Merchants

Time Limit:1000MS     Memory Limit:65536KB     64bit IO Format:%I64d & %I64u

Submit  Status  Practice  HDU 3466

Appoint description:  System Crawler  (Aug 10, 2016 2:34:39 PM)

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
n个商人，每个商人有一个物品，物品有价格w、价值v还有一个交易限制p。p的意义是假如你现在拥有的钱数小于p，那么是不允许交易的。
w是交换时需要给对方的钱，v是评估价值，即自己获得的钱，p是限制条件
因为有条件限制，按背包算之前先排序，贪心选取q小p大的，q-p差值从小到大排序，具体原因不知道-_-|||
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#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int dp[5505];
struct node
{
	int w,p,v;
}a[505];
bool cmp(node x,node y)
{
	return x.p-x.w<y.p-y.w;
}
int main()
{
	int n,m;
	while(~scanf("%d%d",&n,&m))
	{
		memset(dp,0,sizeof(dp));
		for(int i=0;i<n;i++)
			scanf("%d%d%d",&a[i].w,&a[i].p,&a[i].v);
			sort(a,a+n,cmp);
			for(int i=0;i<n;i++)
			{
				for(int j=m;j>=a[i].p;j--)
				{dp[j]=max(dp[j],dp[j-a[i].w]+a[i].v);
				}
			}
			printf("%d\n",dp[m]);
	}
	return 0;
}