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.
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.
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.
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
dp[i][j]表示已经买了i种花费j
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<limits.h>
typedef long long LL;
using namespace std;
int dp[25][220];
int k[25],pri[25][25];
int m,c;
int main()
{
int t;
cin>>t;
while(t--)
{
cin>>m>>c;
for(int i=1;i<=c;i++)
{
cin>>k[i];
for(int j=1;j<=k[i];j++)
cin>>pri[i][j];
}
memset(dp,0,sizeof(dp));
dp[0][0]=1;
for(int i=1;i<=c;i++)
{
for(int j=0;j<=m;j++)
{
if(dp[i-1][j])
{
for(int kk=1;kk<=k[i];kk++)
{
if(j+pri[i][kk]<=m)
dp[i][j+pri[i][kk]]=1;
}
}
}
}
int flag=1;
for(int i=m;i>=0;i--)
{
if(dp[c][i])
{
cout<<i<<endl;
flag=0;
break;
}
}
if(flag) cout<<"no solution"<<endl;
}
return 0;
}