思路：这道题还是用了小小的剪枝，这里要注意的是该题有很多中构建树的顺序，但是，在这众多顺序中不一定都能保证输出的方案字典序最小。

构建搜索树：如图构建

　　　　　

剪枝，emmm，看代码：

#include<iostream>
using namespace std;
int a[20][30];
int b[30], an[30], mm[20], gg[20];
int n, m, mins=1000, kk;
bool f(){
    for (int i = 1; i <= m;++i)
    if (b[i] > an[i])return 0;
    return 1;
}
void add_of_s(int s, int y){
    for (int i = 1; i <= m; ++i)
        an[i] = an[i] + y*a[s][i];
}

void dfs(int s, int k){
    if (k > n){ return; }
    if (kk >= mins){ return; }    //剪枝
    if (f()){  mins = kk; for (int i = 1; i <= mins; ++i)gg[i] = mm[i]; }
    for (int i = s+1; i <= n;++i)
    {  
        kk++;    mm[kk] = i;
        add_of_s(i, 1);
        dfs(i , k + 1);
        mm[kk] = 0; kk--; 
        add_of_s(i, -1);
    }
}

int main(){
    cin >> m;
    for (int i = 1; i <= m; ++i)cin >> b[i];
    cin >> n;
    for (int i = 1; i <= n;++i)
    for (int j = 1; j <= m; ++j)
        cin >> a[i][j];
    dfs(0, 0);
    cout << mins << " ";
    for (int i = 1; i <= mins; ++i)
        cout << gg[i] << " \n"[i == mins];

}