Continuous backpack (Backpack)

[Problem description]
Select some items from the items in the T group and put them in the backpack to find the minimum value of the remaining space.
Restriction: You must select consecutive items from each group. That is to say, if there are n items in this group: item 1, Item 2, Item 3 ,... , Item n, you can only select item I, item I + 1 ,... , Item j, where 1 <= I <= j <= n, or not.
[Input]
The first two positive integers v and T separated by spaces. The number of spaces and items in the backpack.
Next there are T rows. Each row first has a positive integer ni, indicating that this group of items has ni, and then ni positive integers, indicating the size of each item.
[Output]
Only one number indicates the minimum value of the remaining space.
[Input and output sample]
Bag. in
100 3
3 7 6 8
2 80 70
4 101 108 103150
 
Bag. out
6
 
[Example of input and output]
Select 6, 8 for the 1st group, 80 for the 2nd group, and 3rd for the group.
[Restrictions]
60% of data meets the requirements: 1 <= ni <= 10
100% of data meets the requirements: 1 <= ni <=, 1 <= v <=, 1 <= T <= 10
 

For each section of a backpack
Next we will find our backpacks
Note that in a single step Vmax = V [I-1] + sum [ni]

[Cpp]
# Include <cstdio>
# Include <cstring>
# Include <cstdlib>
# Include <iostream>
# Include <cmath>
# Include <functional>
# Include <algorithm>
Using namespace std;
# Define MAXNI (100 + 10)
# Define MAXV (500000 + 10)
# Define MAXN (10 + 10)
Int T, n, v, sum [MAXNI] = {0}, a [MAXNI], A [MAXN] [MAXV] = {0 };
Bool f [MAXN] [MAXV] = {0 };
Int main ()
{
Freopen ("bag. in", "r", stdin );
Freopen ("bag. out", "w", stdout );
Scanf ("% d", & v, & T );
// V = v * 10;
F [0] [0] = 1;
For (int I = 1; I <= T; I ++)
{
Scanf ("% d", & n );
Sum [0] = 0;
For (int j = 1; j <= n; j ++) {scanf ("% d", & a [j]); if (a [j]> v/*/10 */) a [j] = 0; sum [j] = sum [J-1] + a [j];}
// For (int j = 1; j <= n; j ++) printf ("% d", sum [j]);
For (int k = 0; k <= v + sum [n]; k ++)
For (int j = n; j> = 0; j --)
{
If (k-sum [j] <0) continue;
If (f [I-1] [k-sum [j])
{
F [I] [k] = 1;
A [I] [k] = j;
Break;
}
}
For (int k = 0; k <= v + sum [n]; k ++)
For (int j = A [I] [k]-1; j> = 1; j --)
{
F [I] [k-sum [j] = f [I] [k] | f [I] [k-sum [j];
}



// Cout <endl;
}
// V/= 10;
Int I = v;
While (! F [T] [I])
{
I --;
}
Printf ("% d \ n", v-I );
/*
For (int I = 0; I <= T; I ++)
{
For (int j = 0; j <= v; j ++)
If (f [I] [j]) cout <I <',' <j <'';
Cout <'\ n ';
}

While (1 );
*/Return 0;
}