hdu 3033 I love sneakers!

I love sneakers!

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 3340    Accepted Submission(s): 1361

Problem Description

After months of hard working, Iserlohn finally wins awesome amount of scholarship. As a great zealot of sneakers, he decides to spend all his money on them in a sneaker store.

There are several brands of sneakers that Iserlohn wants to collect, such as Air Jordan and Nike Pro. And each brand has released various products. For the reason that Iserlohn is definitely a sneaker-mania, he desires to buy at least one product for each brand.
Although the fixed price of each product has been labeled, Iserlohn sets values for each of them based on his own tendency. With handsome but limited money, he wants to maximize the total value of the shoes he is going to buy. Obviously, as a collector, he won’t buy the same product twice.
Now, Iserlohn needs you to help him find the best solution of his problem, which means to maximize the total value of the products he can buy.

 

Input

Input contains multiple test cases. Each test case begins with three integers 1<=N<=100 representing the total number of products, 1 <= M<= 10000 the money Iserlohn gets, and 1<=K<=10 representing the sneaker brands. The following N lines each represents a product with three positive integers 1<=a<=k, b and c, 0<=b,c<100000, meaning the brand’s number it belongs, the labeled price, and the value of this product. Process to End Of File.

 

Output

For each test case, print an integer which is the maximum total value of the sneakers that Iserlohn purchases. Print "Impossible" if Iserlohn's demands can’t be satisfied.

 

Sample Input

  

   5 10000 3
1 4 6
2 5 7
3 4 99
1 55 77
2 44 66
  

 

Sample Output

  

   255
  

 

Source

2009 Multi-University Training Contest 13 - Host by HIT

 

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题意：给出每一种牌子的不同型号的靴子，每种都可以买或者不买。但是有一个条件，就是每一种牌子最少都需要

一双。

题解：如果没有这限制条件就是简单的01背包问题了。但是加上了限制条件？？？？

其实以后我们解这种题时如果有限制条件就在原来的dp基础上加了一个维，就是加上他的限制条件。。

dp[i][k1] = Max(dp[i][k1],dp[i][k1-node[j].mey]+node[j].val,dp[i-1][k1-node[j].mey]+node[j].val);

dp[i][j]表示挑了1到i种牌子的有j钱可以买的最大价值。

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define INF -0x7fffffff
int dp[11][10002];
int n,m,k;
struct node
{
	int mod;
	int mey;
	int val;
}node[105];
int f[15];
bool cmp(struct node a ,struct node b)
{
	if(a.mod<b.mod)
		return true;
	else
	return false;
	
}
int Max(int a,int b)
{
	return a>b?a:b;
}
int main()
{
	int i,j,k1;
	while(~scanf("%d",&n))
	{
		memset(f,0,sizeof(f));
		scanf("%d%d",&m,&k);
		for(i=0;i<n;i++)
		{
			scanf("%d%d%d",&node[i].mod,&node[i].mey,&node[i].val);
			f[node[i].mod]++;
		}
		for(i=1;i<=k;i++)
			for(j=0;j<=m;j++)
				dp[i][j] = INF;
			for(i=0;i<=m;i++)
				dp[0][i] = 0;
			int flag = 0;
			for(i=1;i<=k;i++)
			{
				if(f[i]==0)
					flag = 1;
				f[i] = f[i-1]+f[i];
			}
			if(flag==1)
			{
				printf("Impossible\n");
				continue;
			}
			sort(node,node+n,cmp);
			/*for(i=0;i<n;i++)
			{
				printf("%d %d %d\n",node[i].mod,node[i].mey,node[i].val);
			}*/
			for(i=1;i<=k;i++)
				for(j=f[i-1];j<f[i];j++)
					for(k1=m;k1>=node[j].mey;k1--)
					{
						dp[i][k1] = Max(dp[i][k1],dp[i][k1-node[j].mey]+node[j].val);
						dp[i][k1] = Max(dp[i][k1],dp[i-1][k1-node[j].mey]+node[j].val);
						
					}
					
					if(dp[k][m]<0)
						printf("Impossible\n");
					else
						printf("%d\n",dp[k][m]);
	}
	return 0;
}