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
     
      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];}