11450 - Wedding shopping

 D. Wedding Shopping 

Background

One of our best friends is getting married and we all are nervous because he is thefirst 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 famousdepartment 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 justone.

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

The Input

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

The Output

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

Sample Input

3100 43 8 6 42 5 104 1 3 3 74 50 14 23 820 33 4 6 82 5 104 1 3 5 55 33 6 4 82 10 64 7 3 1 7

Sample Output

7519no solution

#include<stdio.h>#include<string.h>int a[25][25],f[25][205],b[25],t,m,c,i,j,k;int main(){scanf("%d",&t);while(t--){memset(f,0,sizeof(f));**f=1;scanf("%d%d",&m,&c);for(i=1;i<=c;i++){scanf("%d",&b[i]);for(j=1;j<=b[i];j++)scanf("%d",&a[i][j]);}for(i=1;i<=c;i++)  for(j=0;j<=m;j++)  if(f[i-1][j])  for(k=1;k<=b[i];k++)  if(j+a[i][k]<=m)  f[i][j+a[i][k]]=1;  int max=0;  for(i=m;i>=1;i--)  if(f[c][i]){max=i;break;}  if (max) printf("%d\n",max);  else puts("no solution");  }return 0;}