Wedding shopping - UVa 11450 背包+滚动数组

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朋友即将结婚，一群从未参加过婚礼的人需要购买全套装备。面对有限的预算，如何选择不同衣物的最佳组合以最大化利用资金成为了一个有趣的问题。本文通过一个具体的算法解决这一挑战，确保既不超出预算又能买到所有必需品。

Wedding shopping

Background

One of our best friends is getting married and we all are nervous because he is the first of us who is doing something similar. In fact, we have never assisted to a wedding, so we have no clothes or accessories, and to solve the problem we are going to a famous department store of our city to buy all we need: a shirt, a belt, some shoes, a tie, etcetera.

The Problem

We are offered different models for each class of garment (for example, three shirts, two belts, four shoes, ..). We have to buy one model of each class of garment, and just one.

As our budget is limited, we cannot spend more money than it, but we want to spend the maximum possible. It's possible that we cannot buy one model of each class of garment due to the short amount of money we have.

The Input

The first line of the input contains an integer, N, indicating the number of test cases. For each test case, some lines appear, the first one contains two integers, M and C, separated by blanks (1<=M<=200, and 1<=C<=20), where M is the available amount of money and C is the number of garments you have to buy. Following this line, there are C lines, each one with some integers separated by blanks; in each of these lines the first integer, K (1<=K<=20), indicates the number of different models for each garment and it is followed by K integers indicating the price of each model of that garment.

The Output

For each test case, the output should consist of one integer indicating the maximum amount of money necessary to buy one element of each garment without exceeding the initial amount of money. If there is no solution, you must print "no solution".

Sample Input

3
100 4
3 8 6 4
2 5 10
4 1 3 3 7
4 50 14 23 8
20 3
3 4 6 8
2 5 10
4 1 3 5 5
5 3
3 6 4 8
2 10 6
4 7 3 1 7

Sample Output

75
19
no solution

题意：你手上有M数量的钱，然后你有C种物品需要买，每种都要买一个，如果可以的话，问你最多花费多少钱，否则输出no solution。

思路：滚动数组，数组num1[i]表示用数量为i的钱能不能恰好买到前面所有种类的物品各一样，num2[i]记录加上这种物品之后的状态。

AC代码如下：

#include<cstdio>
#include<cstring>
using namespace std;
int num1[210],num2[210];
int main()
{ int t,n,m,i,j,k,value,ans;
  scanf("%d",&t);
  while(t--)
  { scanf("%d%d",&n,&m);
    memset(num2,0,sizeof(num2));
    num2[0]=1;
    while(m--)
    { scanf("%d",&k);
      for(i=0;i<=n;i++)
      { num1[i]=num2[i];
        num2[i]=0;
      }
      while(k--)
      { scanf("%d",&value);
        for(i=n;i>=value;i--)
         if(num1[i-value]==1)
          num2[i]=1;
      }
    }
    ans=-1;
    for(i=0;i<=n;i++)
     if(num2[i]==1)
      ans=i;
    if(ans==-1)
     printf("no solution\n");
    else
     printf("%d\n",ans);
  }
}